Solution:
Let $s_{1} = \sqrt{x}$, $s_{2} = \sqrt{x + s_{1}}$, $s_{3} = \sqrt{x + s_{2}}$ and so on. So the equation given is $y = s_{1998}$.

We show first that all $s_{n}$ must be integral for $1 \leq n \leq 1998$. $y$ is integral, so $s_{1998}$ is integral. Now suppose $s_{n}$ is integral. Then $s_{n-1} = s_{n}^{2} - x$ is integral, proving the claim.

So in particular $s_{1}$ and $s_{2}$ are integers and $s_{2}^{2} = s_{1}^{2} + s_{1}$. But if $s_{1} > 0$, then $s_{1}^{2} < s_{1}^{2} + s_{1} < (s_{1} + 1)^{2}$, which is impossible. Similarly $s_{1} < 0$ is impossible. So the only possible solution is $s_{1} = 0$ and hence $x = 0$ and $y = 0$.