We are looking for all integer solutions for the equation $a^b+1=b^a$. We conjecture that there are only the solutions $(0,b),(1,2),(2,3)$. It is easy to see, that if a is odd and b even, there is only the solution $(1,2)$, but we don't see the general case. We think there should be a simple argument.

mucheasier than quoting Mihailescu's theorem! $\endgroup$