Prove that $x^2+1$ cannot be a perfect square for any positive integer x? I started this problem by trying proof by contradiction.

I first noted that the problem stated that $x$ had to be a positive integer, and thus $x=0$ could not be a solution. I then assumed that $x^2+1=n^2$ for some integer $n$ other than $1$. From here I have tried various methods, to no avail:


Factoring:


$n^2-x^2=1\implies(n+x)(n-x)=1$. It would be nice if I could say $(n+x)=(n-x)=1$ and $(n+x)=(n-x)=-1$. However, the issue here is that $(n+x)$ and $(n-x)$ could take on any value. For example, $(n+x)=2$ and $(n-x)=\frac{1}{2}$, or $(n+x)=3$ and $(n-x)=\frac{1}{3}$. Thus, I ruled out factoring.


For any number $n$, $n\equiv 0 \pmod4$ or $n\equiv 1 \pmod 4$.


The trouble with this is that I would have to prove the above statement, so I ruled this out.

Does anyone have any tips on how to continue? I feel like this should be an easy proof, but no solutions are coming to me, without having to prove something else.
For any $n$, the square is $n^2$.

The following integer is $n+1$, and its square is $n^2 + 2n + 1$.

Since (assuming $x&gt;0$) $$x^2 &lt; x^2+1 &lt; x^2 + 2x + 1$$ 

it also true that  $$x = \sqrt{x^2}  &lt; \sqrt{x^2+1} &lt; \sqrt{x^2 + 2x + 1} = x+1$$ but there are no integers between $x$ and $x+1$.
You could try proving by contradiction by assuming that $x^2+1$ is a perfect square for some $x∈\mathbb{Z}$ then show how this cannot be possible