This is a solution-verification question.
As a lemma, a group $G$ as in the title is centerless or Abelian. For a proof, see @NickyHekster's answer below or the link in the comments. So, as for the main focus in the title (determining $n_p$ and $n_q$ for such $G$'s), we have to address the following two cases:
- Case "$G$ centerless". If $Z(G)$ is trivial, then every nontrivial element of $G$ has centralizer of order $p$ or $q$. Therefore, the Class Equation yields: $$pq=1+k_pp+k_qq \tag 1$$ where $k_i$ is the number of conjugacy classes of size $i=p,q$. Since $\langle x\rangle=C_G(x)$ for every $x\in G\setminus\{e\}$, there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$); but each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because $q\nmid p-1$). Therefore, $(1)$ yields: $$pq=1+k_pp+m'q(p-1) \tag 2$$ for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields: $$p=n+m'(p-1) \tag 3$$ In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$, and finally $k_q=p-1$). Therefore: $$1+k_pp=q \tag4 $$ Now we have two options: if $p\nmid q-1$, then $(4)$ is false, and hence $G$ can't be centerless (so, it is Abelian, see Case "$G$ Abelian"); on the other hand, if $p\mid q-1$, then by $(4)$ there are exactly $q-1$ elements of order $q$ (namely $\bf{n_q=1}$) and hence $q(p-1)$ elements of order $p$ (namely $\bf{n_p=q}$).
- Case "$G$ Abelian". Suppose there are two distinct cyclic subgroups of order $p$, say $\langle x\rangle$ and $\langle y\rangle$; then, $\langle x\rangle\langle y\rangle$ is a subgroup of $G$ of order $p^2$: contradiction, because $p^2\nmid pq$. So, there is at most one cyclic subgroup of order $p$, namely $n_p\le 1$. By the same argument, there is at most one cyclic subgroup of order $q$ (because $q^2\nmid pq$), namely $n_q\le 1$. Therefore, there are at least $pq-1-(p-1)-(q-1)=$ $(p-1)(q-1)\ge 2$ elements of order $pq$, and for any $x$ of them, $x^q$ has order $p$ and $x^p$ has order $q$, and hence $\bf{n_p=1}$ and $\bf{n_q=1}$. $\space\Box$
