$\rm p\mid pq\mid n^{pq}+1\Leftrightarrow n^{pq}+1\equiv 0\pmod p\Leftrightarrow (n^p)^q\equiv -1\pmod p\Rightarrow $ $\rm \Rightarrow n^{2q}\equiv 1\pmod p\Leftrightarrow ord_p(n)\mid 2q\Leftrightarrow ord_p(n)=1,2,q,2q$ But also $\rm ord_p(n)\not |q$ because $\rm n^q\equiv 1\pmod p $ ,so $\rm ord_p(n)=2,2q$. Similarly we have that $\rm ord_q(n)=2,2p$. If $\rm ord_p(n)=2q,ord_q(n)=2p$ then we have $\rm 2q\leq \phi(p)=p-1,2p\leq \phi(q)=q-1\Rightarrow 2(p+q)\leq p+q-2$,contradiction! So $\rm ord_p(n)=2$ or $\rm ord_q(n)=2$. Without loss of generality suppose that $\rm ord_p(n)=2\Rightarrow p\mid n^2-1$. Let $\rm pq=2f+1$.Then $\rm p\mid n^{pq}+1\Leftrightarrow p\mid n^{2f+1}+1\Leftrightarrow p\mid n+1$. Also we have $\rm p^3q^3\mid n^{pq}+1$ so $\rm p^3|n^{pq}+1$.Let $\rm n^{pq}+1=p^3r$. $\rm LTE\Rightarrow u_p((n^{pq}+1^{pq})=u_p(n+1)+u_p(pq)\Leftrightarrow $ $\rm \Leftrightarrow 3+u_p(r)=1+u_p(n+1)\Rightarrow u_p(n+1)\geq 2\Leftrightarrow p^2|n+1$,qed