My attempt: clearly when $m=1$, $2^{2m+1}+17=25$ is a perfect square. I conjecture that there are only finitely many $m$ such that $2^{2m+1}+17$ is a perfect square, and this might be obtained by considering modulo some value (e.g., 3, 8, etc), but I could not figure out the detail.
On the other hand, this question would be much easier if one looks at $2^{2m}+17$.
Please help.