Note that $2023 = 17^2 \cdot 7$ and let $(x, y)$ be a solution. If we assume that $y$ is not divisible by $17$, there exists a number $a$ such that $a y \equiv x \pmod{17}$. We then have
$$0 = x^2 - x y + y^2 \equiv y^2 (a^2 - a + 1) \pmod{17},$$
hence $a^2 - a + 1 \equiv 0 \pmod{17}$. This implies that $a^6 \equiv 1 \pmod{17}$, because
$$a^6 - 1 = (a^3 - 1)(a^3 + 1) = (a^3 - 1)(a + 1)(a^2 - a + 1).$$
On the other hand, by Fermat's little theorem, $a^{16} \equiv 1 \pmod{17}$. Then $a^2 \equiv a^2 \cdot a^{16} \equiv a^{18} \equiv (a^6)^3 \equiv 1 \pmod{17}$, hence $a \equiv \pm 1 \pmod{17}$. But for these two values of $a$ we get $a^2 - a + 1 \equiv 2 - a \equiv 2 \mp 1 \not\equiv 0 \pmod{17}$, a contradiction. Hence, $y$ must be divisible by $17$.

It follows that $x$ must also be divisible by $17$, and we look for solutions $(x, y) = (17u, 17v)$. Let $N(x, y) = x^2 - x y + y^2$ and note that
$$N(x, y) = \frac{(2x - y)^2 + 3y^2}{4}.$$
Then $2023 = N(17u, 17v) = 17^2 N(u, v)$, iff $7 = N(u, v)$, or equivalently
$$28 = (2u - v)^2 + 3v^2.$$
Because $v^2 \le 28/3$ and $28$ is not a square, to find solutions to this equation we only need to consider $v \in \{\pm 1, \pm 2, \pm 3\}$. To a solution with negative $v$ there corresponds one with positive $v$, obtained by changing the signs of both, $u$ and $v$. Because $28 - 3 = 5^2$, $28 - 12 = 4^2$, and $28 - 27 = 1^2$, we have these three cases for solutions with positive $v$
$$v = 1, \quad 2u - v = \pm 5, \quad v = 2, \quad 2u - v = \pm 4, \quad v = 3, \quad 2u - v = \pm 1.$$
The possibilities for solutions $(u, v)$ with positive $v$ are therefore
$$(3, 1) \quad (-2, 1) \quad (3, 2) \quad (-1, 2) \quad (2, 3) \quad (1, 3).$$
Thus we obtain $12$ solutions to the original equation, namely
$$(x, y) = (17, 51), (34, 51), (-17, 34)$$
and all the variants $(-x, -y)$, $(y, x)$, $(-y, -x)$ of these.