I claim the answer is $k=2$. Example for $k=2$ Taking $x=y$, we find $8x^2$ as Tiranian. So, $16x^2$ is a sum of two Tiranian numbers, and letting $16x^2=6^{4046}$, we find $x=6^{2023}/4$ achieves the equality. Impossibility for $k=1$ We now show $x^2+6xy+y^2=6^{4046}$ is impossible. Suppose the contrary. Considering modulo 3, we find $3\mid x^2+y^2$, so $3\mid x,y$. Continuing, we find $u^2+6uv+v^2 = 2^{4046}$ for some $u,v$. Next, let $d={\rm gcd}(u,v)$. Then, $d=2^t$ for a suitable $t$, and we further deduce $m^2 + 6mn+n^2 = 2^{2t}$ for a suitable $t\ge 1$ and coprime $m,n$; this yields $(m+3n)^2 - 8n^2 = 2^{2t}$. Observe that $m,n$ are thus both odd. Next, if $v_2(m+3n)\ne 2^t$, then it is easy to see that $v_2((m+3n)^2 - 2^{2t})$ is always even, whereas $v_2(8n^2)=3$ is odd, since $n$ is odd, a contradiction. In the last case where $v_2(m+3n)=t$, we find $2^{2t}\mid 8n^2$, and as $n$ is odd, we find $t=1$ as the only possibility. The equation $x^2+6xy+y^2 = 4$, however, admits no solutions. So, $k=1$ does not work.

Proposed by me. I just randomly thought of a quadratic form $f(x,y)$ and wanted to investigate a number being of the form $f(a_1,a_2) + f(b_1,b_2)$ or similar (but of course, wanted something distinct from $f(x,y) = x^2 + y^2$ which is way too well known). It would have been even nicer if the form could be of higher degree, but unfortunately the world is full of (hyper)elliptic curves; anyway, glad that it turned out a good problem like this, hope you like it!

my thought of showing k=1 is not possible is like this m^2+6mn+n^2 is divisible by 6 which implies ,6|m^2+n^2 which gives two possibilities m,n=0(mod6) m.n=3(mod6) from first case we could let m=6p n=6q which gives 36(p^2+q^2+6pq)=6^4046 which is just like previous ,we can make a chain sort of thing by letting it a multiple of 6 and at last we can get x=0 y=6^4046 (NOT POSIBBLE ) CAN ANYONE HELP ME WITH THE OTHER CASE)

$\boxed{ANSWER:K=2}$ Case 1)$k=2$ let's say that $x=y\implies 16x^2=6^{4046}\implies x=y=6^{2023}/4$ Case 2)$k=1$ $x^2+6xy+y^2=6^{2023}\implies 3|x^2+y^2\implies 3|x,3|y$ Let's say $x_1=3x,y_1=3y\implies 9(x_1^2+6x_1y_1+y_1^2)=6^{4046}\implies x_1^2+6x_1y_1+y_1^2=6^{4046}/9$ which is same as before but $9$ times smaller using infinite discent $x_{2023}^2+6x_{2023}y_{2023}+y_{2023}^2=1$ since $x_{2023},y_{2023}>0\implies LHS\geq 8$ contradiction Let's get rid of index so now we have:$x^2+6xy+y^2=2^{4046}$ From case work we get:$x\equiv y\equiv 2\pmod{4}$(1) or $x\equiv y\equiv 0\pmod{4}(2)$ Case 1)$ x=4x_1+2$ and $y=4y_1+2$ we get:$4((2x_1+1)^2+6(2x_1)(2y_1)+(2y_1+1)^2)=2^{4046}$ $2x_1+1\equiv x$ and $2y_1+1\equiv y\implies x^2+6xy+y^2=2^{4044}$ and $x\equiv y\equiv 1 or 3\pmod{4}$ which we know that fails from case work. Case 2)$x\equiv y\equiv 0\pmod{4}$ From now on we have only this case and using infinite descent we finally get: $x^2+6xy+y^2=1$ which is contradiction and we are done