Let $x=m-z$, then we have $p^3 + px = x^2$. Then $p \mid x$, so let $pa=x$, then $p+a=a^2$, so $a=2$ and $p=2$. Therefore $x =4$, or $m=z+4$. Since $m > 0$ and $z<0$, we only have $z \in \lbrace -1,-2,-3 \rbrace$. The solutions are $(2,4+z,z)$ for these $z$.