Suppose that $q>p$ without loss of generality. In particular, if $p\neq 2$ otherwise $8\mid x^2+1$ for a suitable $x$, not possible. Now, let $a=n^p$. We then have $p^3\mid a^q+1$. Thus $a^{2q}\equiv 1\pmod{p}$ and $a^{p-1}\equiv 1\pmod{p}$. Combining, we get $a^{(2q,p-1)}\equiv 1\pmod{p}$. Since $q>p$, $(2q,p-1)\in\{1,2\}$. Moreover, $p\neq 2$, thus $a^2\equiv 1\pmod{p}$ and $a\not\equiv 1\pmod{p}$. Hence, $p\mid n^p+1$. In particular, $a^{q-1}-a^{q-2}+\cdots+1\equiv q\pmod{p}$, and since $p\neq q$, we get one further conclusion: $p^3\mid n^p+1$. From here, lifting the exponent lemma gives $v_p(n+1)\geqslant 2$, as desired.