For $n \ge 2$, we have
$$\frac{a_{n+1} - a_n}{a_n - a_{n-1}} = \frac{4(a_1 + a_2)(a_2 + a_3) \cdots (a_{n-1} + a_n)}{4(a_1 + a_2)(a_2 + a_3) \cdots (a_{n-2} + a_{n-1})} = a_{n-1} + a_n$$
and hence $a_{n+1} = a_n^2 + a_n - a_{n-1}^2$. $a_2 = a_1^2 - 2$ and by induction assuming $a_n = a_{n-1}^2 - 2$ we can show that $a_{n+1} = a_n^2 - 2$.

We get $a_2 = 7 = 45^2 - 2018$, $a_3 = a_2^2 - 2 = 47$, $a_4 = 47^2 - 2 = 2207 = 65^2 - 2018$ and hence $a_n+2018$ is a perfect square for $n=2,4$. Let us show that $a_n+2018$ is not a perfect square for $n>4$. Let $a_k+2018 = c^2$ for some $k > 4$, $c \in \mathbb{Z}^+$. Then $c^2 - a_{k-1}^2 = 2016$. Since $c > a_{k-1}$, we get $2016 = c^2 - a_{k-1}^2 \ge (a_{k-1} + 1)^2 - a_{k-1}^2 = 2a_{k-1} + 1$ and hence $a_{k-1} < 1008$. However, the sequence is increasing and for $k > 4$ we have $a_{k-1} \ge a_4 = 2207 > 1008$ which is a contradiction. The answer is 2.