It suffices to show that $p \mid m$, since then $p \mid n^2 \Longrightarrow p \mid n$. Moreover, the problem is symmetric in $m$ and $n$. So assume now that $p$ divides neither $m$ nor $n$. We have $(m+n)^2-mn = m^2+mn+n^2 \mid m^3-n^3$, hence $m^3 \equiv n^3 \pmod{p}$, and since $p \equiv 2 \pmod{3}$ this implies $m^{p-2} \equiv n^{p-2} \pmod{p}$. But by FLT we also get $m^{p-1} \equiv 1 \equiv n^{p-1} \equiv m^{p-2}n \pmod{p}$. This implies $m \equiv n \pmod {p}$ and consequently $0 \equiv m^2+mn+n^2 \equiv 3m^2 \pmod{p}$, which is contradiction since $p \not= 3$. Thus the proof is complete.

if m,n are not divisible by $\implies$ p$m^2+mn+n^2=(m+2n)^2-3n$ $\implies$ $(\frac{-3}{p})=1$ which is a contradiction by recopricity law