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.

It looks like I need to spell out the details for insipidintegrator.
If $x$ is even, the prime $17$ is the product of $y+2^{x/2}$ and $y-2^{x/2}.$ Averaging, we find $y=9$ whence $x=6.$
If $x$ is odd, write $y-2^{x/2}=\frac{17}{y+2^{x/2}}$. Letting $x=2n+1$, we have
$\Big|\frac{y}{2^n}-\sqrt{2}\Big|=\frac{17}{2^n(y+2^{n+.5})}.$
From Beuker's thesis, If $q=2^k, \ \Big|\frac{p}{q}-\sqrt{2}\Big|>2^{-43.9}q^{-1.8}$ Thus $\Big|p-q\sqrt{2}\Big|>2^{-43.9}q^{.2}$
In our case, $p=y,q=2^n$ so $p+q\sqrt{2}>2q\sqrt{2}.$ Multiplying, $$17=y^2-2^x>2^{-43.9}q^{.2}\cdot 2q\sqrt{2}$$ or $$q^{1.2}<17\cdot2^{42.4}$$
and $$n<38.73955$$ A computer check shows the only solutions are $n=1,2,4.$ These values correspond to $x=3,5,9.$
