More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.
The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.
In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields
such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.
The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and
harmonic generation at
metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.
Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.
The Rydberg formula only works for hydrogenic atoms, and in hydrogenic atoms all the orbitals with the same principal quantum number have (approximately) the same energy. The $2s$ and $2p$ have the same energy as do the $3s$, $3p$ and $3d$, and so on.
The Rydberg formula only works where the potential energy of the electron varies as $r^{-1}$. If we have more than one electron present then the electrons repel each other and they screen each other from the nucleus. As a result the potential is strictly speaking no longer even central, though to a good approximation we can treat the electron potential as central but no longer varying as $r^{-1}$ (more on this here if you're interested).
In hydrogenic atoms the different angular momentum states are only approximately of equal energy because relativistic effects cause a splitting. For example in hydrogen the $2s$ is slightly higher in energy than the $2p$, and this is known as the Lamb shift. However this is a tiny effect.
Best Answer
There are a few reasons why the particle produced needs to be a photon. Aside from conserving energy, we also need to conserve momentum, charge and spin, for example. So you would need to ask what other particle, instead of a photon, could be emitted while satisfying all those conservation requirements.
If you just consider energy and spin conservation, the total amount of energy available in electron transitions in an atom is small, and not enough to make any of the other massive Bosons. To use your terminology, the maximum energy difference in electron transitions, Δ𝐸, is way below the energy Δ𝐸𝑐 you would need to create any of the other known massive particles that satisfy the other conservation requirements.