Notice that $gcd(p , n) > 1$, so $n = p^{\alpha}n_0$ then $p + n \mid pn \Longleftrightarrow$ $p^{\alpha - 1}n_0 + 1 \mid p^{\alpha}n_0$. If $\alpha \geq 2$ $\Longrightarrow gcd(p^{\alpha-1}n_0 + 1, p^{\alpha}) = 1$(false) $\Longrightarrow \alpha = 1$ then $n_0 + 1 \mid pn_0$ $\Longrightarrow n_0 + 1 = p$. $\therefore (n, p) = ((p - 1)p, p)$

Much more direct: $n+p$ divides $p(p+n)-pn=p^2$, hence $p+n \in \{1,p,p^2\}$, but clearly only the third option is possible, hence $n=p^2-p$.