The r.h.s. of the equation is divisible by $3$ but not by $9$. Assume that $3 \mid x$. Then $9 \mid x^3$ and $9 \mid 3xy$. If also $3 \mid y$ then $9 \mid y^3$, implying that the l.h.s. of the equation is divisible by $9$. Thus $3 \nmid y$. But then $3 \nmid y^3$, implying that the l.h.s. of the equation is not divisible by $3$. The contradiction shows that $3 \nmid x$. By symmetry, also $3 \nmid y$.
As $3 \mid 3xy$, we must have $3 \mid x^3 + y^3$. Hence also $3 \mid x^3 + 3x^2y + 3xy^2 + y^3 = (x+y)^3$, implying $3 \mid x+y$. Consequently, $x$ and $y$ are modulo $3$ congruent to $1$ and $2$ in some order. Hence $x^2 \equiv y^2 \equiv 1 \pmod{3}$ and $xy \equiv 2 \pmod{3}$, implying $x^2 - xy + y^2 \equiv 0 \pmod{3}$. We obtain $9 \mid x^3 + y^3 = (x+y)(x^2 - xy + y^2)$, whereas $3xy \equiv 6 \pmod{9}$. Hence the l.h.s. of the equation is congruent to $6$ modulo $9$ but the r.h.s. of the equation is congruent to $3$. Consequently, there are no solutions.

Denote $a = x+y$. Then the given equation is equivalent to $a^3 + 3xy(1-a) = 2019$. As $3 \mid 2019$ and $3 \mid 3xy(1-a)$, we must have $3 \mid a^3$, implying $3 \mid a$. The given equation is also equivalent to
$$(a-1)(a^2 + a + 1 - 3xy) = 2018. \quad (1)$$
Hence $a-1 \mid 2018$. As $2018 = 2 \cdot 1009$ where $1009$ is prime, $a-1$ must be one of $2018$, $1009$, $2$, $1$, $-1$, $-2$, $-1009$, $-2018$. Taking into account that $3 \nmid a$, we obtain four cases:
* If $a = 2019$ then $y = 2019 - x$ and substituting into (1) gives $2019^2 + 2019 + 1 - 3x(2019 - x) = 1$ which is equivalent to $x^2 - 2019x + 673 \cdot 2020 = 0$. The latter equation has no real solutions.
* If $a = 3$ then $y = 3 - x$ and substituting into (1) gives $13 - 3x(3 - x) = 1009$ which is equivalent to $x^2 - 3x - 332 = 0$. The latter equation has no integral solutions.
* If $a = 0$ then $y = -x$ and the l.h.s. of the initial equation is non-positive.
* If $a = -1008$ then $y = -1008 - x$ and substituting into (1) gives $1008^2 - 1008 + 1 + 3x(1008 + x) = -2$ which is equivalent to $x^2 + 1008x + 336 \cdot 1007 + 1 = 0$. The latter equation has no real solutions.