I assume that $x$ and $y$ are integral, in which case your sum is one quarter of the sum
$$\sum_{r=1}^\infty \exp(-\sqrt{r}) s(r),$$ where $s(r)$ is the number of representations of $r$ as the sum of two squares. Clearly, $\sum_{r=1}^R s(r) \sim \pi r^2,$$\sum_{r=1}^R s(r) \sim \pi r,$ which means you can estimate the original sum via summation by parts, followed by approximating the sum by the integral.