The following solution was proposed together with the problem:

The condition is symmetric so we can assume that $b \leq a$ .

The first case is when $a=b$ . In this case, $a!+a=5^m$ for some positive integer $m$ . We can rewrite this as $a \cdot \left ( (a-1)!+1 \right)=5^{m}$ . This means that $a=5^{k}$ for some integer $k \geq 0$ . It is clear that $k$ cannot be $0$ . If $k \geq 2$ , then $(a-1)!+1 = 5^{l}$ for some $l \geq 1$ , but $a-1=5^{k}-1 > 5$ , so $5|(a-1)!$ , which is not possible because $5|(a-1)!+1$ . This means that $k=1$ and $a=5$ . In this case, $5!+5 = 125$ , which gives us the solution $(5,5)$ .

Let us now assume that $1 \leq b<a$ . Let us first assume that $b=1$ . Then $a+1=5^x$ and $a!+1=5^y$ for integers $x,y \geq 1$ . If $x \geq 2$ , then $a=5^x-1 \geq 5^2-1>5$ , so $5|a!$ . However, $5|5^y=a!+1$ , which leads to a contradiction. We conclude that $x=1$ and $a=4$ . From here $a!+b=25$ and $b!+a=5$ , so we get two more solutions: $(1,4)$ and $(4,1)$ .

Now we focus on the case $1 < b < a$ . Then we have $a!+b = 5^x$ for $x \geq 2$ , so $b \cdot \left ( \frac{a!}{b}+1 \right ) = 5^x$ , where $b|a!$ because $a>b$ . Because $b|5^x$ and $b>1$ , we have $b=5^z$ for $z \geq 1$ . If $z \geq 2$ , then $5<b<a$ , so $5|a!$ , which means that $\frac{a!}{b}+1$ cannot be a power of $5$ . We conclude that $z=1$ and $b=5$ . From here $5!+a$ is a power of $5$ , so $5|a$ , but $a>b=5$ , which gives us $a \geq 10$ . However, this would mean that $25|a!$ , $5|b$ and $25 \nmid b$ , which is not possible, because $a!+b=5^x$ and $25|5^x$ .

We conclude that the only solutions are $(1,4)$ , $(4,1)$ and $(5,5)$ .