Let $P$ be the union of the sets of the prime divisors of $a_n$, $n \ge 1$. Denote by $v_p(n)$ the highest power of $p$ that divides $n$. Suppose on the contrary, that $|P|$ is finite. Let $A = \{\prod_{p \in P} p^k : k \in \mathbb{Z}, k > v_p(a_1) \ \forall p \in P\}$. Then $|A| = \infty$. Let $t \in A$. Assume that $a_{t+1} \ne a_1$. Then there exists $q \in P$ such that $v_q(a_{t+1}) \ne v_q(a_1)$. Thus $v_q(a_{t+1} - a_1) = \min(v_q(a_{t+1}), v_q(a_1)) \le v_q(a_1) < v_q(t)$. But this contradicts the fact that $n \mid (a_{t+1} - a_1)$. Thus we must have $a_{t+1} = a_1$. Now for any positive integer $m$, we have $(t+1-m) \mid (a_{t+1} - a_m) = (a_1 - a_m)$. Since this is true for all $t \in A$ and $|A| = \infty$, $a_1 = a_m$ for all $m$, a contradiction.