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: 
$$2^{x}+1=y^{2}\qquad\text{with }\big(x,y\big)\in\mathbb{N}.$$
 A: Hint. If $x,y\in\mathbb{N}$ such that
$$2^x=y^2-1=(y+1)(y-1)$$
then $(y+1)$ and $(y−1)$ are both powers of $2$ and their difference is $2$. What may we conclude? 
A: So $y$ is odd: $y=2z+1$ so $$2^x= 4z(z+1)$$
Since $z$ and $z+1$ are consecutive one must be odd and since only odd factor on left is $1$ we have two possibilities:


*

*$z=1$ then $2^x=8$ so $x=3$ and $y=3$.

*$z+1=1$ then $z=0$ but this can not be.

A: The smallest solution is $x=y=3$. Any larger solution contradicts Mihailescu's theorem, which states $8,\,9$ are the only consecutive integers of the form $n^k$ with $n,\,k\in\Bbb N,\,k\ge2$. But as @RobertZ notes, that's overkill: $y\pm1$ need to be powers of two, making them $2,\,4$, giving the above solution.
A: $$2^{x}+1=y^{2} \implies y=\pm \sqrt {2^x+1}$$
You are interested only in natural numbers solutin so we get 
$$y= \sqrt {2^x+1}$$
For those natural numbers $x$ where $2^x+1$ is a perfect square.
$x=y=3$ is the only solution.
Note that you want to have  $2^x = (y-1)(y+1)$ so both $y-1$ and $y+1$ must be powers of $2$ which is possible only if  $y=3$
A: 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 $x=3$.
