Analyzing a Diophantine equation: $A^k + 1 = B!$ Efficient way to solve.

Given a positive integer $A$, if $k$ and $B$ are positive integers such that $$A^k+1=B!,$$ it is clear that $A^k$ and $B!$ are coprime. Then also $A$ and $B!$ are coprime, and so $B$ is strictly smaller than the smallest prime factor of $A$. If $A$ is not too large, an effective approach is to determine the smallest prime factor of $A$, and then simply try all values of $B$ up to that prime. In particular, for your example with $A=99$ we see that the smallest prime factor is $3$, so we only need to try $B=2$ to see that there are no solutions.

Note that if you intend to test this for many values of $A$, it may be worth while to verify that $B!-1$ is not a perfect power for any small value of $B$. (Thanks to Peter in the comments $B!-1$ is not a perfect power if $B\leq10^4$.)

Some more general results: A quick check shows that every solution with $B\leq3$ is of the form $$(A,k,B)=(1,k,2)\qquad\text{ or }\qquad(A,k,B)=(5,1,3).$$ For $B\geq4$ we have $A^k=B!-1\equiv7\pmod{8}$ and so $k$ is odd and $A\equiv7\pmod{8}$. Then $$B!=A^k+1=(A+1)(A^{k-1}-A^{k-2}+A^{k-3}-\ldots+A^2-A+1),$$ which shows that $A+1$ divides $B!$, so in particular $A+1$ is $B$-smooth. So for every prime $p$ dividing $A$ and every prime $q$ dividing $A+1$ we have $q\leq B<p$.