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.
 A: 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 $x=3,5,9.$
A: 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.

A: A largely complete answer:
Starting from $2^x+17=y^2$, we obtain $2^x+2^4=(y-1)(y+1)$. By observation, $x=4$ fails as $33\ne y^2$. Note that $(y-1),(y+1)$ are two consecutive even numbers, so their difference is $2$, and one of them contains a single factor of $2$ while the other contains more than one factor of $2$.
Case 1, $x<4 \Rightarrow x=1,2,3$: Hence, $2^x+2^4=18,20,24$. Of these, only $24=4\times 6$ is the product of two consecutive even numbers. This corresponds to $x=3$.
Case 2, $x>4$: $16(2^{x-4}+1)=(y-1)(y+1)$. We see $(2^{x-4}+1)$ is an odd number, which we can factor as $ab$, where both $a,b$ are odd integers $\ge 1$. It must be the case that $(y-1)(y+1)=8a\cdot 2b$ where $8a-2b=\pm 2$, or $4a-b=\pm 1 \Rightarrow b=4a\pm 1$ and $(2^{x-4}+1)=ab=4a^2\pm a$
For $a=1$ we get $4a^2\pm a=3,5$ corresponding to $2^1+1$ and $2^2+1$, whence $x-4=1,2 \Rightarrow x=5,6$
For $a=3$ we get $4a^2\pm a=33,39$. Of these, $33=2^5+1 \Rightarrow x-4=5 \Rightarrow x=9$
I note that for $x\ge 9 \Rightarrow x-4 \ge 5$ it is the case that $(2^{x-4}+1) \equiv 1 \bmod 32$ and $a$ being odd means $a^2 \equiv 1 \bmod 8 \Rightarrow 4a^2 \equiv 4 \bmod 32$. Thus $(2^{x-4}+1)=4a^2\pm a \Rightarrow 4 \pm a \equiv 1 \bmod 32$, or $a\equiv \pm 3 \bmod 32$. Looking at the first $200$ members of $\{3,29,35,61,67,\dots\}$ I find no further suitable values of $a$
For larger $a$, I have not found a proof that none exist. If I find such a proof, I will update this answer.
A: 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!
A: $x=3,5,9$ work with $y=5,7,23$ but fail nor $x=1,7$.  If $x $ works for any other value then $x>4$.
$2^x+17=y^2$
$2^4 (2^{x-4}+1)=y^2-1=(y+1)(y-1) $
$y\pm 1$ must be even and $\gcd(x+1,x-1)=2$ and one is a multiple of $4$ while the other isn't.
That means::
$y\pm 1=8m;y\mp 1=2n;mn=2^{x-4}+1$
Also means 
$2^x+8=y^2-9$
$2^3 (2^{x-3}+1)=(y-3)(y+3) $
So $y\mp 3 =4k=2n\mp 2;y\pm 3=2j=8m\pm 2;jk=2^{x-3}+1$
We have strict limitation here it seems.  In that $m,n $ are relatively close to $j,k $ but their product is about half.
I'm too lazy to work out the details but there must be an upper limit to that being possible.
Also $2^x-8=2^3 (2^{x-3}-1)=y^2-25=(y-5)(y+5) $ and so $y\pm 2 =4r;y\mp 2=2s;rs=jk-2$
So.... it's a mess but we have clear limits.  
I strongly suspect $x=9$ is the highest possible $x $.
Oh, and we have $2^x+1=(y-4)(y+4)\approx 16mn $.  
