I claim the answer is $m=2^k$, $k\ge 1$ is arbitrary. The given recursion rearranges to $a_{n+1}-1=(n+1)(a_n-1)$; thus, a telescoping argument shows that $a_n = 2\cdot n!+1$ for $n\ge 0$. Now $a_n$ is odd for all $n$. We next show that for any $p\ge 3$ prime, there is an $n$ such that $p\mid a_n$. The case $p=3$ is handled by $n=0$, so assume $p\ge 5$. Note that by Wilson's theorem, $(p-1)!\equiv -1\pmod{p}$, and therefore $(p-3)!\equiv -1/2\pmod{p}$. Consequently, $a_{p-3}\equiv 0\pmod{p}$.