Let $z=0,y=1,x=1+t$. $P_n|Q_n$ implies $f(t)=(1+t)^{2n}+t^{2n}+(t^2+t)^{2n}$ divides $g(t)=((1+t)^{2n}+1+t^{2n})^{2n}$. Since they are monic polynomials, the quotient should have all integer coefficients. Then $f(1)|g(1)$, $2^{2n+1}+1|(2^{2n}+2)^{2n}$, or $2^{2n+1}+1|3^{2n}$. So $2^{2n+1}+1=3^k$ for some $k$. It is easy to show the only solution is $n=1$. $P_1|Q_1$ indeed!

We could also use Zsigmondy... er, I mean, the result of #4... to get that if $n>1$, some prime divides $2^{2n+1}+1$ that doesn't divide $2^{2n-1}+1$

How would that work? Place #4 in front of 1?

By Gauss's Lemma, $ \frac{Q_n(x, y, z)}{P_n(x, y, z)} \in \mathbb{Q}[x, y, z] \Longrightarrow \frac{Q_n(x, y, z)}{P_n(x, y, z)} \in \mathbb{Z}[x, y, z] $. Letting $ (x, y, z) = (1, 2, 3) $ this implies $ 2^{2n + 1} + 1\vert (2^{2n} + 2)^{2n} \Longrightarrow 2^{2n + 1} + 1\vert (2^{2n + 1} + 4)^{2n} \Longrightarrow 2^{2n + 1} + 1\vert 3^{2n} $. Now, Zsigmondy destroys this immediately, but since #4 literally says "prove Zsigmondy" I feel like that's cheating. Instead, by LTE, we have that $ v_3(2^{2n + 1} + 1) = 1 + v_3(2n + 1) $ so it suffices to show that $ 3^{1 + v_3(2n + 1)} < 2^{2n + 1} + 1 $ which will imply that $ 2^{2n + 1}+ 1 $ is not a power of three. But note that $ 3^{1 + v_3(2n + 1)} \le 6n + 3 < 2^{2n + 1} + 1 $ as desired when $ n > 1 $. Therefore the only $ n $ that works is $ \boxed{n = 1} $.

We see that $n = 1$ works. We prove that no $n > 1$ fits. Plug in $z = 0$ and $y = 1$. We get \begin{align*} P_n(x, 1, 0) = (x-1)^{2n} + x^{2n} + (x-1)^{2n}x^{2n} \end{align*}and \begin{align*} Q_n(x, 1, 0) = [(x-1)^{2n} + x^{2n} + 1]^{2n}. \end{align*}Let $x$ be a root of $P_n(x, 1, 0)$. By the assumption now $Q_n(x, 1, 0) = 0$. This results in \begin{align*} (x-1)^{2n} + x^{2n} = -1 \end{align*}and \begin{align*} (x-1)^{2n}x^{2n} = 1. \end{align*}The system $a+b=-1, ab = 1$ has the solution $\{a, b\} = \{\zeta_3, \zeta_3^2\}$, where $\zeta_k$ is a primitive $k$th of unity. It follows that $x$ is a $6n$th root of unity, and so is $x-1$. This is possible if and only if $x$ is $\pm \zeta_6$. Hence $P_n(x, 1, 0)$ has only the numbers $\pm \zeta_6$ as its roots. They must have equal multiplicity as $P_n(x, 1, 0)$ has real coefficients. Thus \begin{align*} (x-1)^{2n} + x^{2n} + (x-1)^{2n}x^{2n} = (x^2 - x + 1)^{2n}. \end{align*}Plug in $x = 2$ to get $1 + 2^{2n+1} = 3^{2n}$. This implies $n = 1$ by size reasons, a contradiction.