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 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.
