Prove that $x^2+1$ cannot be a perfect square for any positive integer x?

For any $n$, the square is $n^2$.

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

Since (assuming $x>0$) $$x^2 < x^2+1 < x^2 + 2x + 1$$

it also true that $$x = \sqrt{x^2} < \sqrt{x^2+1} < \sqrt{x^2 + 2x + 1} = x+1$$ but there are no integers between $x$ and $x+1$.

Your factoring is abosolutely correct. However, since $x,n$ are integers, this gives us $x-n$ is a natural number. Thus $x-n = \pm 1$.

An alternate proof would use that a square can not exist between two consecutive squres and $$x^2 < n^2 < x^2+2x+1$$