Pencil of polynomials mostly irreducible?

Let $p,q \in \mathbb{Q}[x]$ two relatively prime polynomials. Let $h\in \mathbb{R}$ any number and let $F_h(x) = p(x) + h \cdot q(x)$.

What can be said about the irreducibility of the polynomial $F_h(x)$ over the field $\mathbb{Q}(h)$?

More precisely, let $H = \{h : F_h \mbox{ is reducible over }\mathbb{Q}(h) \}$. How big can $H$ be?

Is there some reasonable condition on $p$ and $q$ that guarantees that $H$ is empty? Or finite? Or at least discrete in some sense?

I understand that $H$ can only contain algebraic numbers. The linearity of $F$ in $h$ forces $F$ to be irreducible for transcendental $h$, because one of the factors would have to lie in $\mathbb{Q}(x)$ and that would contradict that $p$ and $q$ are coprime.

I would guess that the same is true when $h$ is algebraic but its degree is large relative to the degrees of $p$ and $q$, though I don't know how to approach this.

Surely, it is worth assuming $\max(\deg p,\deg q)>1$ (and $\deg q\geq \deg p$).
In this case, there is no hope for $H$ to be finite or discrete. Take any positive integer $n$ and any nonzero rational $r$. Let $h$ be a root of $G_{r,n}(x)=p(rx^{2n})+xq(rx^{2n})$ (this polynomial is of odd degree!). Then $h\in H$, since $x-rh^{2n}\mid F_h(x)$.
Fixing $n$ and varying $r$, you get that $H$ is dense in the set of value of $(-p/q)$.
Moreover, I would expect the polynomial $G_{r,n}(x)$ to be irreducible infinitely often; in this case $h$ can have arbitrarily high degree.