$|G|<+\infty$, $p\not=q$ primes, $n_p=p+1$, $n_q=q+1$.
I want to show that $\exists P\in Syl_p(G), Q\in Syl_q(G)$ s.t. $PQ=P\times Q<G.$
I understand this exercise as to show that $PQ$ is a subgroup.
If one of tham is normal, then the statement is trivial. However, neither $n_p$ nor $n_q$ is $1$, we don't know the normalities of $P$ and $Q$.
The main problem is showing the closedness of the operation: $a=p_1q_1, b=p_2q_2.$ $ab=p_1q_1p_2q_2\in PQ?$
From the Sylow theorems, I've got some facts like $$|G|=p^nq^mA\implies p+1|q^mA,\ q+1|p^nA$$ but how to connect them to the closedness.
You must assume $\{p,q\} \neq \{2,3\}$ (see counterexample of Jyrki below - $S_4$).
Suppose $p \lt q$, let $P \in Syl_p(G)$. By assumption $|G : N_G(P)| = p + 1$. If $q \mid (p + 1)$, then $q \gt p$ yields $q = p + 1$ and $\{ p, q \} = \{ 2, 3 \}$, contradiction.
Hence $q$ does not divide $p + 1$ and $N_G(P)$ contains a full Sylow $q$-subgroup $Q$ of $G$. Thus $H = PQ$ is a subgroup of $G$, with $P \lhd H$. The number of Sylow $q$-subgroups of $H$ is $\equiv 1$ mod $q$ and at most the number in $G$. Thus either $Q \lhd H$ or $H$ has $q + 1$ Sylow $q$-subgroups.
In the latter case, since $|H : Q| = |P|$ is a power of $p$, we obtain $q+1 = p^a$, so $q = p^a −1 = (p−1)(p^{a-1} + \cdots +1)$. Since $q$ is prime, $p \neq 2$ and $q \gt p$, this is a contradiction. Thus both $P$ and $Q$ are normal in $H$, whence $H=P \times Q$.