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.
I think that there are not solutions for $x$ odd.

Indeed,

$y^2=2^x+17=2^{2k+1}+17=2 \cdot 4^k +17 \equiv 2+17 \equiv3 \mod4$, which cannot happen.

(If $x$ is of the form $x=2k$, then the equation can be easily written $(y-2^k)(y+2^k)=17$, and so we go on).

EDIT: Wrong solution!