Can a capacitor with a fixed plate overlap area and separation have a maximum amount of charge it can hold? Can a capacitor with a finite plate area and plates' separation have a saturation point of charge it stores, assuming that the dielectric does not break down? If there is, what formula can be used to find that saturation point? If there isn't, does it then mean that a capacitor's plate-surface can hold an infinite amount of charge? What makes that possible?
 A: According to basic capacitor theory there is no limit to the amount of charge, $±Q$ on the plates nor, therefore, on the electric field, $E=Q/\epsilon_0A$,  between the plates.
As you imply, for capacitors with a dielectric, even just air, there is in practice, an  upper limit to $Q$, imposed by the breakdown field for the dielectric.
But suppose that there is a vacuum between the plates. In that case the upper limit is imposed by the phenomenon of cold field emission. Under very high electric fields, in the order of $1\ \text{GV m}^{-1}$, electrons could 'tunnel' out of the negative plate and be emitted from it, crossing the gap.
Using $E=Q/\epsilon_0A$ we see that a field strength of $1\ \text{GV m}^{-1}$ corresponds to a surface charge density of about $10^{-2}\ \text{C m}^{-2}$, that is about $6 \times 10^{16}\  \text{excess electrons m}^{-2}$. A crude estimate of the surface density of atoms in a copper surface is $2 \times 10^{19}\  \text{atoms m}^{-2}$, so the number of excess electrons can nowhere near exceed the number of surface atoms before cold field emission takes place.
[Note also that the field strength of $1\ \text{GV m}^{-1}$ might be reached at the edges of the plates or at small surface bumps of  high curvature, even though the 'official' field strength of $Q/\epsilon_0 A$ is less than this.]
