Let $\mathbb{Z}^2$ be the set of integer squares. \[2n^2-9mn+m^4=0 \implies (9m)^2-4(2)(m^4)=36m^2-8m^4=m^2(36-8m^2) \in \mathbb{Z}^2 \]Thus, $36-8m^2>0 \implies m=\pm 1$ or $ \pm2$. Since $36-8=24 \notin \mathbb{Z}^2$, $m=\pm 2$. Thus, substituting $m$ and solving the quadratic for $n$, we get $2n^2-18n+16=0\implies (m,n)=(2,1)$ or $(2,8)$ and $2n^2+18n+16=0\implies (m,n)=(-2,-1)$ or $(-2,-8)$.