Prove that if $m^p+n^p\equiv0\pmod p$ then $m^p+n^p\equiv0\pmod {p^2}$ where $p$ is an odd prime number.

You are aware that $ m \equiv - n \pmod {p}$. Let $m = -n + pk$, then

$$m^p + n^p = (pk - n)^p + n^p = \\ (pk)^p + {p\choose 1 } (pk)^{p-1}(- n)^1 + \ldots + { p \choose p-1 } (pk)^{1} (-n)^{p-1} + {p \choose p } (pk)^ 0 (-n)^p + n^p. $$

When $p$ is odd, notice that the last 2 terms cancel out.
Can you show that the rest of the terms are multiples of $p^2$?
Use your observation that ${ p \choose k } $ is also a multiple of $p$.

We have $m+n\equiv 0\pmod p$ which means $m+n=cp$ for some integer $c.$ Then $$ m^p+n^p=(cp-n)^p+n^p =\sum\limits_{k=0}^p \binom{p}{k} (cp)^k(-n)^{p-k} +n^p\\ =(-n)^p + \binom{p}{1} (cp)^1(-n)^{p-1} +\sum\limits_{k=2}^p \binom{p}{k} (cp)^k(-n)^{p-k} +n^p \\ = p^2 c(-n)^{p-1} +\sum\limits_{k=0}^{p-2} \binom{p}{k+2} (cp)^{k+2}(-n)^{p-k-2} \\ = p^2 c(-n)^{p-1} + p^2 c^2 \sum\limits_{k=0}^{p-2} \binom{p}{k+2} (cp)^k(-n)^{p-k-2} $$