= Solution
The curves have no common component, since their intersection is finite. Let $R=0$ define the irreducible degree-$m$ curve $E$. Choose $p\in E$ outside $C\cap D$, and select $(\lambda,\mu)\ne(0,0)$ with $\lambda P(p)+\mu Q(p)=0$. The polynomial $H=\lambda P+\mu Q$ is not identically zero, since $P,Q$ are not proportional.
The degree-$n$ curve $H=0$ meets $E$ at the prescribed $nm$ base points and also at $p$. If $R$ did not divide $H$, irreducibility of $R$ and the <Bézout theorem> would bound the number of distinct intersections by $nm$, contradicting these $nm+1$ points. Thus
$$
H=RS,\qquad\deg S=n-m.
$$
At every remaining common zero of $P,Q$ lying off $E$, $R$ is nonzero and hence $S=0$. Therefore
$$
\boxed{F=\{S=0\}\text{ contains all remaining }n(n-m)\text{ points},\qquad\deg F\leq n-m.}
$$
This is <residual intersection through a curve pencil>. The existence of a point $p$ outside the finite base locus is automatic on the projective curve $E$.
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