Since $a_0 = 0 = a_m$, we can prove by induction that $a_k = a_{m+k}$ for all $k$. If there are two consecutive terms of $\{a_n\}$ which are the same, then all subsequent terms are the same. By the periodicity, all terms are the same, and they are equal to $a_0 = 0$. This gives $a_1 = 0$ and we are done.

Now, assume $d_n = a_{n+1} - a_n \ne 0$ for all $n$. Using the fact $x - y \mid P(x) - P(y)$ for all $x, y \in \mathbb{Z}$, we know that $d_n$ divides $P(a_{n+1}) - P(a_n) = a_{n+2} - a_{n+1} = d_{n+1}$ for all $n$. This gives
$$|a_1 - a_0| = |d_0| \le |d_1| \le |d_2| \le \dots \le |d_m| = |a_{m+1} - a_m| = |a_1 - a_0|.$$

Therefore, all equalities hold, and we have $|d_n| = |d_0|$ for all $n$, which means $d_n = \pm d_0$ for all $n$. Now, note that
$$d_0 + d_1 + \cdots + d_{m-1} = a_m - a_0 = 0.$$
Since $d_n \neq 0$, there are both positive and negative terms among $d_0, d_1, \dots, d_{m-1}$. WLOG assume $d_k$ and $d_{k+1}$ have different signs for some $k < m$. Then
$$a_{k+2} = a_k + d_k + d_{k+1} = a_k \pm d_0 \mp d_0 = a_k.$$
This implies $a_{n+2} = a_n$ for all $n \ge k$, and so $a_2 = a_{m+2} = a_m = 0$ as desired.