Let us show that $(3,2)$ is the only solution. To begin with, note that $a \mid b^a-b^2-1$ which is odd, hence $a$ is odd. Also note that clearly $a$ and $b$ must be coprime and that $(1,1)$ is not a solution. Hence $a \ne b$. If $b>a$, then $a^b>a^a \ge a!$ and hence $b^a<b^2+ab+1 \le 2b^2 \le b^3$ and hence $a=1$, but then $b+1=b^2+b+2$ which is clearly absurd. Hence we must have $b<a$ and $a \ge 2$. Considering the equation modulo $2$, we see that $b$ must be even. Let $b=2^ku$ with $u$ odd. Then the LHS is congruent to $1 \pmod{2^{k+1}}$ hence $2^{k+1} \mid a!+ab$. Hence $2^{k+1} \nmid a!$. But $b<a$ implies that $2^{k+1} \mid a!$ unless $b=2$. But then $4 \nmid a!$ implies that $a=3$ leading to our only solution.