Finding solutions to $2^x+17=y^2$ 
Find all positive integer solutions $(x,y)$ of the following equation: 
  $$2^x+17=y^2.$$

If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is odd.

The only solutions are $x = 3, 5, 6, 9$. This is proved on pp. 148-152 of

Tzanakis, N. (1983), On the diophantine equation $y^2-D = 2^k$, J. Number Theory 17, 144-64.

