Solve in integers

Question

Solve in integers

$2^x = y^2 + 1$

So, functions are generally different. And we can see that $x,y = (1,1),(1,-1),(0,0)$

Answer 1

Clearly $y$ is odd for $x\ge1$

and $x\ge0$ to keep $y$ an integer

So, $y^2\equiv1\pmod4$

Now $2^x-1\equiv-1\not\equiv1\pmod4$ for $x\ge2$

Answer 2

Modulo $4$, $y^2\equiv0 $ or $1$, so $y^2+1\equiv 1$ or $2$.

But for $x\ge2$, $2^x\equiv 0\pmod 4$.

So if $2^x= y^2+1$ then $x<2$.

Can you take it from here?