Why is the number of isotopes of an element bounded? Is there a known reason why any given element has finitely many isotopes? Here I mean both stable and unstable isotopes.
If we know this, do we have a reason why, for a given element, are the isotopes limited to that particular number?
 A: In the comments you clarify your question by asking, as an example, about hydrogen isotopes.
If you look at a Table of Nuclides, you will see that there are at least 7 hydrogen isotopes which have been identified so far. There are links attached to each entry in the table that give data on the reactions for creating these exotic nuclides.
You can see that the He-3 through He-10 have been identified, C-8 through C-22. At the extremely neutron-rich ends, the halflives are extremely short, and neutron emission is prevalent.
It seems, based on the experimental data, that the only restriction on neutron-rich isotopes of any element are the experimental ability to make the nucleus long enough/in sufficient number to get repeatable measurements to demonstrate they have actually been made.
EDIT: As pointed out in a comment, there most likely is a limiting halflife. Check out this question and answer.
There's no reason to assume that H-8 and H-9 won't eventually be formed and identified. It will take a lot of ingenuity, patience, and money.
A: I think this is a good question -- after all, if there's no extra Coulomb repulsion penalty for adding more neutrons, unlike for protons, why can't nuclei have lots of neutrons?
One model for the nucleus we use is called the Semi-Empirical Mass Formula (SEMF), which has a bunch of terms describing the energy contributions to the nucleus.  See wikipedia for the full formula.  The main term that answers your question is the "Asymmetry Term", given by
$$a_\text{A}\frac{(N-Z)^2}{A}$$
where $a_\text{A}$ is some constant we can find empirically, $N$ is the number of neutrons, $Z$ is the number of protons and $A=N+Z$ is the nucleon number.
This is a penalty term in the energy of a nucleus.  If there is a large difference in $N$ and $Z$, this term is large.  If $N$ is similar to $Z$, the term is not as large.  The rational for this is the Pauli Exclusion Principle, which tells us identical particles cannot occupy the same energy state.  If we're adding lots of identical neutrons, we must put them in different energy states.  We can get a cheaper energy cost by filling in some protons instead for a given nucleon number $A$.
To answer your question in the comments:  why do isotopes often have more neutrons than protons, I think the answer there is it is somewhat favourable to add nucleons to the nucleus, because that increases the strong force present, but its cheaper to use neutrons than protons, at least for cases where the ratio $N/Z$ does not deviate too far from 1.
