Since $x^2 + p^2y^2 \equiv 0 \pmod{3}$, if $p \ne 3$ then $x^2 + y^2 \equiv 0 \pmod{3}$, and consequently $x \equiv y \equiv 0 \pmod{3}$. Then $9 \mid x^2 - 3xy + py^2$ but $9 \nmid 12$, a contradiction. Thus, $p = 3$ and $x^2 - 3xy + 9y^2 = 36$. Therefore, $3 \mid x$. $x = 3k$, and we get a second order equation $k^2 - ky + y^2 - 4 = 0$. The discriminant $\Delta = 16 - 3y^2 = m^2$. Therefore, $y^2 = 0$ or $y^2 = 4$. Solutions are: $(-6, 0, 3)$, $(0, -2, 3)$, $(0, 2, 3)$, $(6, -2, 3)$, $(6, 0, 3)$, $(6, 2, 3)$.