Since $2^{2018} \equiv 1\pmod{n}$, we have that $$2^n\equiv 2^r\pmod{n},$$where $r$ is the residue of $n$ when taken modulo $2018$. Furthermore, since $1009$ is prime, we have that by FLT $$n \equiv \frac{2^{2018} - 1}{3}\equiv \frac{2^{2018 - 2\cdot 1009} - 1}{3} \equiv \frac{2^2 - 1}{3}\equiv 1\pmod{1009},$$and combined with the fact that $$n \equiv \frac{2^{2018} - 1}{3}\equiv \frac{-1}{3}\equiv 1\pmod{2},$$we obtain by CRT that $n\equiv 1\pmod{2018}$, implying $r = 1$, and so $$2^n \equiv 2^r\equiv 2\pmod{n}\iff n \mid 2^n - 2,$$as desired.