The sequence is given by the formula $a_n = 2 \cdot n! + 1$ for $n \ge 0$. (We use the usual definition $0! = 1$, which satisfies $1! = 1 \cdot 0!$, in the same way we have $n! = n \cdot (n-1)!$ for other positive integers $n$.) We will prove the equality by induction. We have $a_0 = 3$, which equals $2 \cdot 0! + 1$. Now suppose for certain $k \ge 0$ that $a_k = 2 \cdot k! + 1$, then
$$a_{k+1} = a_k + k(a_k - 1) = 2 \cdot k! + 1 + k \cdot 2 \cdot k! = 2 \cdot k! \cdot (1+k) + 1 = 2 \cdot (k+1)! + 1.$$
This finishes the induction.

We see that $a_n$ is always odd, hence $\gcd(2, a_n) = 1$ for all $n$. It follows also that $\gcd(2^i, a_n) = 1$ for all $i \ge 1$. Hence, $m = 2^i$ with $i \ge 1$ satisfies the condition. Now consider an $m \ge 2$ which is not a power of two. Then $m$ has an odd prime divisor, say $p$. We will show that $p$ is a divisor of $a_{p-3}$. By Wilson's theorem, we have $(p-1)! \equiv -1 \pmod{p}$. Hence,
$$\begin{aligned} 2 \cdot (p-3)! &\equiv 2 \cdot (p-1)! \cdot ((p-2)(p-1))^{-1} \\ &\equiv 2 \cdot -1 \cdot (-2 \cdot -1)^{-1} \equiv 2 \cdot -1 \cdot 2^{-1} \equiv -1 \pmod{p}. \end{aligned}$$
So indeed we have $a_{p-3} = 2 \cdot (p-3)! + 1 \equiv 0 \pmod{p}$. We conclude that $m$ does not satisfy the condition. Hence, the only values of $m$ satisfying the condition are powers of two. $\square$