This was instasolve. The idea is you consider a $n$-step staircase for $2(n+2).$ This has $\frac{n(n+1)}{2}$ unit squares. You can fit two $n$-step staircases to make a $n$ by $n+1$ rectangle, so you can put $2n(n+1)$ staircases as a square. Then, you also split the $\frac{n(n+1)}{2}$ unit squares into smaller squares, so a square and $2(n+2)$ gon are compatible. Then, its also not hard to prove if $P,Q$ and $Q,R$ are compatible, so are $P,R.$

A really really trash writeup although its pretty easy to solve. We start off with some definitions. Let a $n-$ladder be a polymino with $n$ sides consisting of $\frac{n}{4}$ dominoes, stacked next to each other, in the style of a staircase. Let a $n-$M tile be a polymino consisting of a $(n-2)-$ladder with a single monomino attached to the right of the top-most domino of the ladder. Further, a $n-$ladder or M-tile is called vertical, if the dominoes which make up the polymino are oriented vertically. Else, we say that it is horizontal. For example, a $12-$ladder and a $14-$M tile, both of which are vertical, are given below. [asy][asy] import geometry; size(8cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; filldraw((path)((0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,4)--(2,4)--(2,3)--(1,3)--(1,2)--(0,2)--cycle), white+0.1*pri, pri); filldraw((path)((5,0)--(6,0)--(6,1)--(7,1)--(7,2)--(8,2)--(8,3)--(9,3)--(9,4)--(7,4)--(7,3)--(6,3)--(6,2)--(5,2)--cycle), white+0.1*pri, pri); [/asy][/asy] Now, we prove this via induction on $m$. Let a pair $(m,n)$ be called good if there exists two compatible polygons with $m$ and $n$ sides. Now, it is clear that all pairs $(n,n)$ are good (simply consider a $n-$ladder or $n-$M tile (based on $n\pmod{4}$) which can be partitioned into 1 polygon congruent to itself). Now, we have two cases. Case 1 : $n \equiv 0 \pmod{4}$. Here, we place 2 $n-$ladders one horizontally and the other vertically such that their jagged sides coincide. For example, when $n=12$, [asy][asy] import geometry; size(5cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; filldraw((path)((0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,4)--(2,4)--(2,3)--(1,3)--(1,2)--(0,2)--cycle), white+0.1*pri, pri); filldraw((path)((0,-1)--(2,-1)--(2,0)--(3,0)--(3,1)--(4,1)--(4,2)--(3,2)--(2,2)--(2,1)--(1,1)--(1,0)--(0,0)--cycle), white+0.1*tri, tri); [/asy][/asy] Note that this resulting figure will have $n+2$ sides since the $n-4$ step sides do not add to the number of sides (they are present in the normal $n-$ladder similarly) and of the other 4 sides, 2 do not add to the side count since 1 each overlaps with the left-down most side and the bottom side of the $n-$ladder. In particular, in the above figure, the new resulting polygon has only $14$ sides as desired. Thus, in this case we have that $(n+2,n)$ is good. Now, we for $m = n +4k \equiv 0 \pmod{4}$, we simply take $k+1$ $n-$ladders and stack them side to side, with their jagged sides perfectly fitting into each other. For example, when $n=16$, [asy][asy] import geometry; size(4cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; filldraw((path)((0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(4,3)--(4,5)--(3,5)--(3,4)--(2,4)--(2,3)--(1,3)--(1,2)--(0,2)--cycle), white+0.1*pri, pri); filldraw((path)((-1,1)--(0,1)--(0,2)--(1,2)--(1,3)--(2,3)--(2,4)--(3,4)--(3,6)--(2,6)--(2,5)--(1,5)--(1,4)--(0,4)--(0,3)--(-1,3)--cycle), white+0.1*sec, sec); filldraw((path)((-2,2)--(-1,2)--(-1,3)--(0,3)--(0,4)--(1,4)--(1,5)--(2,5)--(2,7)--(1,7)--(1,6)--(0,6)--(0,5)--(-1,5)--(-1,4)--(-2,4)--cycle), white+0.1*tri, tri); label("$\ddots$",(-1,5)--(-1,6)); [/asy][/asy] Each new $n-$ladder clearly contributes exactly 4 new sides. Then, for $m \equiv 2 \pmod{4}$, we simply take the construction for $m=n+2$, and adjoin the stack of $n-$ladders on to the top-left corner of it like above. This again works since each new $n-$ladder will contribute exactly 4 new sides. Case 2 : $n\equiv 2 \pmod{4}$. Here, we place two $n-$M tiles one horizontally and the other vertically, just like before such that their jagged sides coincide. For example, when $n=10$, [asy][asy] import geometry; size(5cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; filldraw((path)((0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(2,3)--(1,3)--(1,2)--(0,2)--cycle), white+0.1*pri, pri); filldraw((path)((0,-1)--(2,-1)--(2,0)--(3,0)--(3,1)--(3,2)--(2,2)--(2,1)--(1,1)--(1,0)--(0,0)--cycle), white+0.1*tri, tri); [/asy][/asy] Note that this resulting figure will have $n-2$ sides since the $n-4$ step sides do not contribute to the number of sides. The other 4 do not either since the left and right sides align perfectly creating no new sides. Thus, 2 sides of the $n-$M tile will now be closed in addition, resulting in $n-2$ total sides as desired. For example, the above polygon has only $8$ sides, as expected. Here, we have that $(n-2,n)$ is good. Now, we for $m = n +4k - 2 \equiv 2 \pmod{4}$, we simply take $k+1$ $n-$M-tiles and stack them side to side, with their jagged sides perfectly fitting into each other. For example, when $n=14$, [asy][asy] import geometry; size(4cm); defaultpen(fontsize(9pt)); pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; filldraw((path)((0,0)--(1,0)--(1,1)--(2,1)--(2,2)--(3,2)--(3,3)--(4,3)--(4,4)--(3,4)--(2,4)--(2,3)--(1,3)--(1,2)--(0,2)--cycle), white+0.1*pri, pri); filldraw((path)((-1,1)--(0,1)--(0,2)--(1,2)--(1,3)--(2,3)--(2,4)--(3,4)--(3,5)--(2,5)--(1,5)--(1,4)--(0,4)--(0,3)--(-1,3)--cycle), white+0.1*sec, sec); filldraw((path)((-2,2)--(-1,2)--(-1,3)--(0,3)--(0,4)--(1,4)--(1,5)--(2,5)--(2,6)--(1,6)--(0,6)--(0,5)--(-1,5)--(-1,4)--(-2,4)--cycle), white+0.1*tri, tri); label("$\ddots$",(-1,5)--(-1,6)); [/asy][/asy] Each new $n-$M tile clearly contributes exactly 4 new sides. Then, for $m \equiv 0 \pmod{4}$, we simply take the construction for $m=n-2$, and adjoin the stack of $n-$lM-tiles on to the top-left corner of it like above. This again works since each new $n-$M-tile will contribute exactly 4 new sides. Thus, it follows that for all even $m,n$ , $(m,n)$ is a good pair, finishing the problem.