Proposed by me, with the idea to be an easy P1. Clearly $x=0$ if and only if $y=0$; suppose without loss of generality that $x>0$ and hence $y>0$. Write $x^2 = y(2021-5y)$ (so that $1\leq y \leq 404$) and note that the gcd of the multipliers on the right divides $2021 = 43 \cdot 47$. If it is $2021$, then $y$ is divisible by $2021$, impossible. If it is $43$ (similarly with $47$), then $y=43z$ gives that $z(47-5z)$ is a perfect square and a direct check (or noticing that $z$ and $47-5z$ must also be squares and then using modulo $5$) shows that there is no solution. If it is $1$, then with $y=t^2$ we get $m^2 + 5t^2 = 2021$ for some positive $m$ and $t$ and now a quick brute force (which can be reduced by using e.g. mod $3$ and mod $5$) concludes that all solutions are $(0,0)$, $(\pm 390, 100)$ and $(\pm 408, 289)$.