Given that we have squares $\left(n^{6}\right.$ and $\left.m^{2}\right)$, we look at the equation modulo 4. To look at it modulo 4, suppose $k \geqslant 2$, the equation becomes $n^{6} \equiv m^{2}+2(\bmod 4)$. Since a square is either 0 or 1, the left-hand side is 0 or 1, and the right-hand side is 2 or 3.

It remains to handle the case $k=1$: for $k=0, (n, m)=(1,1)$ is a solution. The only $k$ that works is $k=1$.
Find all integers $k \geqslant 1$ such that there exists a pair of integers $(n, m)$ such that $9 n^{6}=2^{k}+$ $5 m^{2}+2$.