Square both sides to get $2x+60+2\sqrt{x^2+60x}=y$. So $\sqrt{x^2+60}$ is an integer. Let $x^2+60=m^2$ to get $$(x+30)^2-m^2=(x+30+m)(x+30-m)=900.$$To make $x$ largest, we need to maximize $$x+30=\frac12(x+30+m+x+30-m).$$Since their product is fixed, their sum reaches maximum when they're the farthest from each other, that is $\begin{cases}x+30+m=450,\\x+30-m=2.\end{cases}$In this case, $x$ reaches maximum $196$.

Problem from the future? Quote: Source: 2023 Israel National Olympiad P2