Let $n = 2016$. Let $p_1, p_2, \cdots, p_n$ be $n$ distinct primes bigger than $n$. Consider the system of congruences: \begin{align*} a &\equiv p_1-1 \pmod{p_1^2} \\a &\equiv p_2-2 \pmod{p_2^2} \\ & \cdots\\a &\equiv p_n-n \pmod{p_n^2} \end{align*}Take a solution $a$ of these system of congruences (it exists due to the Chinese Remainder Theorem), and we may assume this $a$ is positive. Let $a_i = a+i$, then $p_i | a_i$, but $p_i^2 \not | a_i$. Also, note that $a_i$ is the only term that $p_i$ divides, since if $p_i | a_j$ then $p_i | a_j -a_i = (j-i)$, so $j = i$ since $p_i > n > |j-i|$. To finish, let the product of $a_i$ be $P = a_1 \cdot a_2 \cdots a_n$, and let $b_i = Pa_i = aP + iP$. Take any $j \neq i$, then $p_j | b_i$ but $p_j^2 \not | b_i$, therefore $b_i$ is not a perfect power. Also, the product of $b_i$ is $P^n \cdot P = P^{n+1}$ is a perfect power. So $\{b_i \}$ is an arithmetic progression which satisfies the desired conditions.