Bulgarian Winter Tournament

We will first show that $k=1$ is not possible, i.e. $a^2+6ab+b^2=2^{4048}\cdot 3^{4048}$ has no solution in natural numbers. If such $a,b$ exist, then $a^2+b^2$ is divisible by $3$ and therefore $a$ and $b$ are divisible by $3$. Writing $a=3a_1$, $b=3b_1$ and dividing by $3^2$, we get $a_1^2+6a_1{b}_1+b_1^2=2^{4048}\cdot 3^{4046}$; repeating this argument another $2023$ times, we arrive at an equation of the form $$u^2+6\nu \upsilon +{\upsilon }^2=2^{4048}$$ where $\nu$ and $\upsilon$ are natural numbers.

If $\upsilon$ is even, then $\nu$ is even; writing $\nu =2u_1$, $\upsilon =2v_1$ and dividing by $4$, we get $u_1^2+6u_1{v}_1+v_1^2=2^{4046}$; repeating this several times, we arrive at an equation of the form $$s^2+6st+t^2=2^{2A},$$ where $s,t,A$ are natural numbers, $t$ is odd and $A\ge q\ge 2$ (since the left side is greater than or equal to $8$). The latter is equivalent to $(s+3t{)}^2-8t^2=2^{2A}$. Now from modulo $8$ we see that $(s+3t{)}^2$ is divisible by $8$, so $s+3t$ is divisible by $4$, i.e. $(s+3t{)}^2$ is divisible by $16$. But then $8t^2$ must be divisible by $16$, which is impossible for an odd $t$, a contradiction. Therefore, $k=1$ is not possible.

For an example with $k=2$ let us first notice that $$36=[1^2+6\cdot 1\cdot 1+1^2]+[3^2+6\cdot 3\cdot 1+1^2]$$ and now multiplying by $(6^{2023}{)}^2$ leads to $$36^{2024}=[(6^{2023}{)}^2+6\cdot 6^{2023}\cdot 6^{2023}+(6^{2023}{)}^2]+[(3\cdot 6^{2023}{)}^2+6\cdot (3\cdot 6^{2023})\cdot 6^{2023}+(6^{2023}{)}^2]$$ i.e. $36^{2024}$ is the sum of the numbers $a_1^2+6a_1{b}_1+b_1^2$ and $a_2^2+6a_2{b}_2+b_2^2$, where $a_1=b_1=b_2=6^{2023}$ and $a_2=3\cdot 6^{2023}$.

$\square$