$p\mid (n+1)^3-1=(n)(n^2+3n+3)\implies p\mid n$ or $n^2+3n+3$. But since $n\mid p-3$ and $p>3$, we have $p-3\geq n$, so $p\nmid n$ as otherwise, $n\geq p$. Thus, $p\mid n^2+3n+3$. Let $p=nk+3$. We have \begin{align*} nk+3\mid n^2+3n+3&\implies nk+3\mid (n^2+3n+3)(k)-(nk+3)(n) \\ &\implies nk+3\mid 3nk+3k-3n \\ &\implies nk+3\mid nk+k-n \\ &\implies nk+3\mid nk+k-n-(nk+3)\\ &\implies nk+3\mid k-n-3 \\ \end{align*}Assume for the sake of contradiction that $k-n-3\neq 0$. Then $|k-n-3|\geq |nk-3|$ which is easy to see that it’s impossible. Thus, $k-n-3=0$ which gives the desired result.