It is clear that $x$ and $y$ must be either both even numbers, or both odd.

It is obvious that $3|x$. Let $x=3x_1\implies 3x_1^2 + y^2 = 1998x_1$.

Now we have $3|y$. Let $y=3y_1 \implies x_1^2 + 3y_1^2 = 666x_1$.

Repeat this process, we find the original $x$ and $y$ are both multiples of $27$.

Let $x=27a$ and $y=27b \implies a^2 + 3b^2 = 74a$. Rearranging: $$(a-37)^2 + 3b^2=37^2$$

By applying the conclusion in <myArtical>12</myArtical>, we find: $$\left\{\begin{align}|a-37|&= p^2 - 3q^2\\b &= 2pq\\37&=p^2+3q^2\end{align}\right.$$

Starting from the last relationship: $$q^2< \frac{37}{3}\implies q = 1, 2, 3\implies (p, q) = (5, 2)$$

Hence, the solutions to the original equation are: $$(a, b) = (24, 20), (50, 20)\implies (x,y)=(648, 540), (1350, 540)$$