Bulgaria competition problems, 2022

We will prove that it's not possible. Let $k = \sqrt{xy + x + y}$ and $z = x + y + 2k + 1$. We have
$$y < \sqrt{y^2 + 2y} \le \sqrt{xy + x + y} = k \le \sqrt{x^2 + 2x} < x + 1.$$
Hence $k \in (y, x]$ and $z \in (x + 3y + 1, 3x + y + 1]$. It's easy to check that $xy + yz + zx = (x + y + k)^2$ and $xy + yz + zx + x + y + z = (x + y + k + 1)^2$. Let assume that there exists a prime number $p$, such that $p\mid(x + y + z, x^2 + xy + y^2)$. Then $x + y \equiv -z \pmod p$ and so $x^2 + xy + y^2 = (x + y)^2 - xy \equiv -xy - yz - zx \pmod p$. Hence $p\mid xy + yz + zx$ and $p\mid x + y + z$, and therefore $p\mid(x + y + k)^2$ and $p\mid(x + y + k + 1)^2$, which is impossible.