Can't be too big, can it?

Note that since $\text{gcd}(18,25)=1$ , $18 \mid y^2$ . Thus, $6 \mid y$ , so the minimum value of $y$ is 6. Substituting, we get $18x = 25 \times 6^2$ , so $18x = 900$ , which implies $x=50$ is the minimum value.