Answer: all odd numbers.
For $n = 1$ there is a solution $3 \cdot 27^2 - 13^2 = 2018^1$. For any odd number $n = 2k+1$ from the latter equality we can obtain the equality $3 \cdot (27 \cdot 2018^k)^2 - (13 \cdot 2018^k)^2 = 2018^{2k+1}$, which means that all odd $n$ satisfy the conditions of the problem.

If $n$ is even, $2018^n \equiv 2^n \equiv 1 \pmod{3}$. But the left side of this equality cannot be equal to 1 modulo 3, since the perfect squares are congruent to 0 or 1 modulo 3. Hence, there are no even numbers $n$ satisfying the conditions of the problem.