$(a_1, a_2)$ can be $(1, -2015), (15, -69), (70, -14), (2016, 0)$ or their permutations.

Firstly, taking the difference of the two relations $a_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$ and $a_{n+1} + a_{n+2} = 2a_{n+3}a_{n+4} + 2016$, we obtain
$$a_{n+2} - a_n = 2a_{n+3}(a_{n+4} - a_{n+2}).$$
Inductively, this easily implies
$$a_{n+2} - a_n = 2^k a_{n+3} a_{n+5} \cdots a_{n+2k+1} (a_{n+2k+2} - a_{n+2k})$$
for any $k \in \mathbb{Z}^+$. This shows $2^k \mid a_{n+2} - a_n$ for any $k \in \mathbb{Z}^+$, and hence $a_{n+2} = a_n$. Thus, we may assume all odd terms are equal to $a_1 = b$ and all even terms are equal to $a_2 = c$. Now the only condition is
$$b + c = 2bc + 2016.$$
This means
$$(2b - 1)(2c - 1) = 2(2bc - b - c) + 1 = -4031 = -29 \times 139.$$
Thus, we obtain the solutions $(b, c) = (1, -2015), (15, -69), (70, -14), (2016, 0)$ up to permutation.