How can I solve $2^{x}+1=y^{2}$? I found an exponential equation with two variables $x$ and $y$, and I can't find the solutions. I want to solve it for $x$ and $y$. I tried using logarithms but I couldn't solve it. This is the equation: 
$$2^{x}+1=y^{2}\qquad\text{with }\big(x,y\big)\in\mathbb{N}.$$

Observe that $2^x=y^2−1=(y+1)(y−1) \Rightarrow y+1 = 2^a$ and $y-1 = 2^b$, where $x=a+b$. Then, we have $a \geq b$, and $2^a-2^b= (y+1) - (y-1) = 2$, which gives us $2^b(2^{a-b}-1)=2$. So, we have the following cases: 
(I) $\left\{\begin{array}[rcl]  
22^b& = & 1 \\
2^{a-b}-1 & = & 2
\end{array} \right. $ $\quad$ (II) $\left\{\begin{array}[rcl]  
22^b& = & 2 \\
2^{a-b}-1 & = & 1
\end{array} \right. $
Note that (I) is not possible. From (II), $b=1$ and $a=2$, and so $y=3$ and $x=3$.
