Solving Diophantine Equation

Question

$x$,$y$ are integers such that $~x^{2}+1=y^{x}$.
Find all pairs of $(x,y)$.

I know that it's a diophantine equation but don't have any idea.
Also I can't find anything related to it by search function.

Answer 1

Well, y = $\sqrt[x]{x^2+1}$. How many integers of this form are there? This function has a maximum value (~2.2) and a minimum (~0.4), so you can bound $y \in [1,2]$. If $y=1$, the equation is $x^2+1=1$, which is trivial. If $y=2$, the equation is $x^2+1 = 2^x$ and it's easy to show via the derivative that there are no solutions with $x>5$ (RHS too big). Since the LHS is an integer for all integer $x$ and the RHS is not an integer for $x<0$ you have another bound. Searching this space is pretty quick...