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 $a \cdot ((a-1)! + 1) = 5^m$. 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 $5 \mid (a-1)!$, which is not possible because $5 \mid (a-1)! + 1$. 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 $1 \leq b < a$. We have $a! + b = 5^x$ and $b! + a = 5^y$. Since $b < a$, $b! + a < a! + a$, so $5^y < 5^x$, which implies $y < x$. If $x \geq 2$, then $5 \mid a!$. However, $5 \mid 5^y = a! + 1$, 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 $1 < b < a$. Because $b \mid 5^x$ and $b > 1$, we have $b = 5^z$ for $z \geq 1$. If $z \geq 2$, then $5b = 5$, which gives us $a \geq 10$. However, this would mean that $25 \mid a!$, $5 \mid b$ and $25 \nmid b$, which is not possible, because $a! + b = 5^x$ and $25 \mid 5^x$.

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