If $x=p^{3}$, $y=p^{5}-p$, then $\frac{x^{3}-y}{1+xy}=\frac{p^{9}-p^{5}+p}{1+p^{8}-p^{4}}=p$ and $0<x<y$ for $p\geq 2$? so $p\geq 2$ are good numbers. $p=0$ is also good (for example, $x=2, y=8$). If $p\leq-1$, then $y-x^{3}\geq 1+xy$ $x>xy$ - contradiction. If $p=1$, then $x^{3}-1=(x+1)y$ $x+1\mid x^{3}-1$ and $x+1\mid 2$, because $x+1\mid x^{3}+1$ $x=1$, hence $y=0$ - contradiction with beeing $y$ positive. So, answer is $p\geq 2$ and $p=0$.