solving a Diophantine equation $17+2^m=n^2$

Question

What is the method to solve $$17+2^m=n^2$$ for positive integers $m$ and $n$. I tried using modular arithmetic which seems did not help.

Answer 1

As a hint: Take $m$ odd/even like below $$(1): \ m=2k \to 17+2^{2k}=n^2\\17=n^2-2^{2k}\\17=17*1 \to \\17*1=(n-2^k)(n+2^k)\\\begin{cases}n+2^k=17 \\n-2^k=1\end{cases}$$ for the odd $m $ $$(2): \ 17+2^{2k+1}=n^2\\2^{2k+1}+1=n^2-16\\2^{2k+1}+1\underbrace{\equiv}_{3} 0$$ so RHS must be multiple of $3$ or $$n^2-16 \underbrace{\equiv}_{3} 0$$