Let $n$ be an integer, then $$n\equiv 0, \pm 1, \pm 2, \pm 3\pmod{7}\implies n^3\equiv 0, \pm 1\pmod{7}$$

Therefore, the sum of two cubes can only be $$x^3+y^3\equiv 0, \pm 1, \pm 2\pmod{7}$$

However, $2019\equiv 3\pmod{7}$. This means the given equation has no solution.