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 $a_1^2 + 6a_1b_1 + b_1^2 = 2^{4048} \cdot 3^{4046}$; 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 $u_1^2 + 6u_1v_1 + v_1^2 = 2^{4046}$; 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 $A \geq q \geq 2$ (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 $a_1^2 + 6a_1b_1 + b_1^2$ and $a_2^2 + 6a_2b_2 + b_2^2$, where $a_1 = b_1 = b_2 = 6^{2023}$ and $a_2 = 3 \cdot 6^{2023}$.

$\square$