Answer: $(p, a, m) = (2, 1, 1), (2, 3, 2), (2, 7, 3), (3, 1, 1), (3, 7, 2), (3, 25, 3), (5, 1, 2), (2, 15, 4), (5, 101, 3)$.
Direct checks for $p = 2, 3, 5$ give the above solutions.
For $p \ge 7$ the Wilson theorem implies $a \equiv 1 \pmod p$. Moreover, obviously $p-1|a-1$. Therefore $a = kp(p-1)+1$ for some integer $k \ge 0$. If $k \ge 6$ then $a \ge 6p^2 - 6p + 1 > 5p^2$, a contradiction. So we have $0 \le k \le 5$.
After division by $p-1$ we obtain
$$(p-2)! + kp = p^{m-1} + p^{m-2} + \dots + 1.$$
Since $p \ge 7$, the $2$ and $\frac{p-1}{2}$ are different and appear in $(p-2)!$, i.e. $p-1|(p-2)!$. This gives $k \equiv m \pmod{p-1}$.
If $m \ge p$, then
$$(p-2)! + kp = p^{m-1} + p^{m-2} + \dots + 1 > p^{p-1} > (p-1)!,$$
which easily gives a contradiction. Thus $m \le p-1$.
If $k=0$, then $m = p-1$ and
$$(p-2)! = p^m - 1 \ge p^{p-1} - 1 > (p-1)^{p-1} > (p-1)!,$$
a contradiction. If $k=1$ or $2$, then $m=1$ or $2$, respectively, which is impossible. If $k=3$, then $m=3$, i.e. $(p-2)! = (p-1)^2$, which is impossible.
For $k=4$ we get $m=4$ and $(p-2)! = (p-1)(p^2+2p-1)$, whence
$$p-2|p^2+2p-1 = p^2-4+2(p-2)+7,$$
i.e. $p=7$, which does not give solutions. For $k=5$ we obtain $m=5$ and $(p-2)! = (p-1)(p^3+2p^2+3p-1)$, whence
$$p-2|p^3+2p^2+3p-1 = p^3-8+2(p^2-4)+3(p-2)+21,$$
i.e. $p=7$, which does not give solutions.