Solve $2^x+7=y^2$ for integer $(x,y)$ How many ordered solutions $(x,y)$ are there to the equation $2^x+7=y^2$ , where $x$ and $y$ are integers?  

I tried taking cases for $x$ and $y$ like they are even or odd but I couldn't solve further. 


Given: $2^x+7=y^2$
Since $L.H.S.$ is odd, $R.H.S$. must be odd.
Putting $y=2m+1$,
$2^x+6=4m^2+4m$,
or $2^{x-1}+3=2m^2+2m$
this forces $x=1$
putting it in original equation, we get $y=3$ and $y=-3$
Thus we have solutions $(1,3)$ and $(1,-3)$
