Inspired by IMO shortlist question

Let $x=zx'$ and $y=zy'$ . Then the equation turns into $z^3x'^3+z^2y'^2+z=z^3x'y'$ or $z^2x'^3+zy'^2+1=z^2x'y'$ Since the RHS is divisible by $z$ it follows that the LHS is also divisible by $z$ , so $z\mid 1\implies z=1$ and our original equation reduces to $x^3+y^2+1=xy\implies y^2-xy+(x^3+1)=0$

Since $y$ is a real number, the discriminant of the quadratic w.r.t $y$ must be positive, i.e $x^2-4(x^3+1)\ge 0\implies x^2\ge 4x^3+4$ which clearly cannot be satisfied if $x$ is a positive integer; thus, there are no solutions.