Dispose of the cases $a = 1, 2, 4$ with the only solution $(4, 1)$. Henceforth $a+1 > 5$We must have $a+1 \mid a!+1$ which by Wilson's theorem, forces $a+1$ to be a prime, say $p$. We want to solve the equation $(p-1)!+1 = p^{2b}$. Rearrange the terms and divide by $p-1$ to obtain that $p^{2b-1}+p^{2b-2}+\dots+1 = (p-2)!$. Read both sides modulo $p-1$ to obtain that $p-1 \mid 2b-1$, which implies $2b \geqslant p$. But then $(p-1)!+1 = p^{2b} > p^p$, a contradiction. Hence the only solution is $(4, 1)$.

This problem is easily solvable for any power $n$ and not just $2^b$! Indeed, $a+1 = p$ is prime and in $(p-1)! = p^n -1$ dividing by $p-1$ and taking the resulting equation mod $p-1$ shows $p-1 \mid n$ if $p\geq 5$ and trivial bounding says bye bye! A related more general problem by me is from Bulgaria National Round 2017 P4.

Assassino9931 wrote: This problem is easily solvable for any power $n$ and not just $2^b$! Indeed, $a+1 = p$ is prime and in $(p-1)! = p^n -1$ dividing by $p-1$ and taking the resulting equation mod $p-1$ shows $p-1 \mid n$ if $p\geq 5$ and trivial bounding says bye bye! Yes the proof supplied by me does not use $2^b$ in any substantial way either.