If we multiply through by $pab$, we find that for $a, b \ne 0$, our equation is equivalent to $pb+pa=ab$, which we can rewrite as
$$(a-p)(b-p)=p^2.$$
We are looking for non-zero solutions of this equation. Conveniently, $p^2$ does not have many factorizations!
We can have $a-p=-1$, $b-p=-p^2$, which yields a negative $b$, or $a-p=-p^2$, $b-p=-1$, which yields a negative $a$.
We could have $a-p=-p$, $b-p=-p$, but that yields the impossible $a=b=0$.
We can have $a-p=1$, $b-p=p^2$, or $a-p=p^2$, $b-p=1$, which respectively yield the solutions $a=p+1$, $b=p^2+p$, and $a=p^2+p$, $b=p+1$.
Finally, we can have $a-p=p$, $b-p=p$, which yields $a=b=2p$.
Comment: The same idea can be used to find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n$ is a given positive integer. The factorizations of $n^2$, with the exception of $n^2=(-n)(-n)$, supply the integer solutions.
Best Answer
We first show that $a \leq b$.
Assume by contradiction that $a >b$. Then
$$a^b=b!( a(a-1)...(a-b+1) +1) \,.$$
Thus $a(a-1)...(a-b+1) +1 | a^b$. But $gcd(a,a(a-1)...(a-b+1) +1)=1$ thus $gcd(a^b,a(a-1)...(a-b+1) +1)=1$. This shows that $a(a-1)...(a-b+1) +1=1$ which is not possible since $a>b$.
We now prove that $a \leq 2$.
Since $a \leq b$ than $a!| a!+b!=a^b$.
Assume by contradiction that $a > 2$. Then, by Bertrand Postulate, there exists a prime $p$ so that $p|a!$ and $p$ doesn't divide $a$. Hence $p|a!$ but $p$ doesn't divide $a^b$, contradiction.
We proved that $a=2$, now the rest is easy.