Determine all pairs $(n, k)$ of non-negative integers such that

$$
2023+2^{n}=k^{2}
$$
Since we have squares and a power of 2, it is natural to look modulo 4. Thus, if $n \geqslant 2$, the equation $2023+2^{n}=k^{2}$ modulo 4 becomes $2023 \equiv 3 \equiv k^{2}(\bmod 4)$, which is impossible because the only squares modulo 4 are 0 and 1.

Thus we must have $n=0$ or 1. If $n=1$, the equation becomes $k^{2}=2023+2=2025=45^{2}$, so $k=45$ is the unique solution: thus we have the pair $(1,45)$ as a solution. If $n=0$, the equation becomes $k^{2}=2024$. But $44^{2}<2024<45^{2}$, so 2024 is not a square: in this case there is no solution.

Thus the unique solution pair is $(n, k)=(1,45)$.