Belarusian Mathematical Olympiad

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, $2018^n\equiv 2^n\equiv 1(\mathrm{mod}3)$. 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.