I claim that for each natural $n \equiv 1\pmod{2018}$, $2018$ divides $a_n$. This is enough to prove the desired statements (as there are infinitely naturals that are $1\pmod{2018}$). To show this, simply note that by FLT, $$a_n \equiv 2\underbrace{000\cdots 0}_{n+2 \text{ 0's}} + 18 \equiv 2\cdot 10^{n+2} + 18 \equiv 2\cdot 10^{(n+2) \pmod{1008}} + 18 \equiv 2\cdot 10^{1+2} + 18 \equiv 2018 \equiv 0\pmod{1009},$$and combined with the fact that $2 \mid a_n \iff a_n \equiv 0\pmod{2}$ for all $n$ (as $a_n$ ends with $8$), we have that since $2$ and $1009$ are relatively prime, $2018 \mid a_n$, as desired.