Does gravity limit the number of bosons that can occupy the same single-particle state? QFT says that an unlimited number of bosons can occupy the same "state" (what I mean by that is that the whole system's wavefunction is composed of a product of many identical wavefunctions).
However, gravity increases monotonically with energy density.  It seems that at some point, one additional boson would create a high enough energy density to create a black hole.  Is this true?  Could I calculate the number of bosons necessary to cause this?
 A: Yes, and Yes.
So how would you go about doing this? Let's start Newtonian. Assume everything is in a ground state. You assume a spherically symmetric wave function. For each shell of radius r, the mass enclosed is the total mass multiplied by the probability of the particle being inside r, which requires integrating the square of the wave function. From the density you can calculate how the energy changes with depth (again, involving integrals). You need to solve for these simultaneously because the wave function depends on the gravitational field and visa-versa.
As long as things stay newtonian it's stable; if you reduce the linear size by 1/2 the gravitational well deepens by a factor of two (gravitational potential drops off like 1/x, force is 1/x^2) but the quantum "zero point" energy increases 4 fold (the particle in a box energy). This means that if the star expands, gravity wins and pulls it back, and if it shrinks, the zero-point term wins and keeps it form collapsing.
Einstein makes things much more difficult. The Schodinger equation no longer applies, and there is not a simple relativistic analogue, and we have to now apply it in a curved space. Finally, we need to know both pressure and density is a function of "circumferential radius". However, we still can use the Newtonian calculations to get an order-of-magnitude estimate of when the whole thing collapses to a black hole.
