If you have more neutrons than protons, then there will be more strong force present to counteract the repulsive forces between protons. Why is it that above bismuth, no nucleus is stable, regardless of its N:Z ratio?
The answers by Alex and userTLK are correct but incomplete.
It is true that whilst the strong force essentially only acts between nearest neighbours, whilst coulomb repulsion acts between all protons, it is actually the weak force that prevents the building of extremely large nuclei.
For example, one must explain why you can't build nuclei with more and more neutrons, simply increasing the neutron to proton ratio.
The answer is that neutrons decay (via a weak interaction) into protons (and electrons) providing there is a spare quantum state for the proton to drop into. If this is not the case then the beta decay is "blocked" by the Pauli exclusion principle. Thus highly neutron rich nuclei will be unstable to beta decay.
Interestingly, in a neutron star crust, beta decay can also be blocked by surrounding degenerate electrons (if their Fermi energy is high enough). There, you can build enormous (atomic mass more than 300) neutron-rich nuclei.
The reason is because the strong force isn't cumulative but the electromagnetic force is. Now, the strong force is a bit more complicated as it does change based on the number of protons and neutrons, but it doesn't build continuously as more protons or neutrons are bound to the nucleus, but the electromagnetic force does.
Say you have a helium atom, 2 protons, 2 neutrons, each is tightly bound by the strong force and the 2 protons are only repelled only by each other. So, it's 1 strong force of attraction, and 1 electromagnetic force of repulsion and the strong force wins. The strong force is 137 times stronger
Now, take Uranium, 92 protons. Each Proton and Neutron is bound to the nucleus by the strong force, but it's only 1 strong force attraction but each proton is now repulsed by 91 other positively charge protons. Hence you have 91 little forces pushing it away. This is much less stable.
Quantum instability always happens at 83 or more protons (1-82 are mostly stable, except for Technetium and Promethium, with 43 and 61 protons respectively), which in and of itself is rather curious. The strong force binds tighter with specific combinations, and as a general rule, even numbers of protons are more often stable than odd numbers. I'm not sure why that is but it seems consistently true.
More on that here: https://en.wikipedia.org/wiki/Even_and_odd_atomic_nuclei and, loosely related, here: http://io9.com/the-oddo-harkins-rule-shows-the-universe-hates-the-odd-1446581327
Also, please see Rob Jeffries answer, as he mentions the weak force - I think his answer is more correct than mine.
Although there are more neutrons than protons to counteract the electrostatic repulsion, there still is proton repulsion. This repulsion grows with larger and larger atoms. By emitting alpha radiation or helium nuclei, an atom can transition from a high energy state to a lower energy state. This is why it is favored as a decay. The more protons the more repulsion and higher energy state.
It's similar to electrons which go from higher energy states to lower ones by emitting photons. The more energetic the electron the more likely it is to emit a photon and transition down to a lower energy. Nuclei are similar but under the influence of different forces.
As the mass number increases, the repulsive effect becomes more significant, because the strong nuclear force (that binds all the nucleons) is very short range, while Coulomb force is not: that is why energy tied up in repulsive Coulomb force bound between protons increases more rapidly as the mass numbers increases than energy tied up in attractive strong nuclear force bound between any two nucleons.
That's why heavy nuclei become unstable.