Suppose there is a 50 watt infrared bulb and a 50 watt UV bulb. Do they emit light of the same energy?

The infrared (IR) bulb will emit more photons per second than the ultraviolet (UV) bulb because each IR photon has less energy than a UV photon.

After some follow-up questions, I've realized that there is an ambibuity in the phrase "50-watt bulb." Is this a bulb that emits 50-watts of light, or a bulb that consumes 50 watts of electrical power? If the former, then my first paragraph is still true. However, most light bulbs are labeled based on the latter. So, in the case of a 50-watt IR bulb and a 50-watt UV bulb, it matters how the light is generated.

For example, if both bulbs are incandescents (the electrical power heats up a filament to thousands of degrees to emit light) with filters in front of them that only let through a desired spectrum, then the IR light power emitted will be much greater than the UV power emitted. This is because the filament is a black body radiator that emits much more power in the lower energy (IR) part of the spectrum. A UV blacklight works by taking a light bulb and putting a filter in front of it that absorbs visible and lower energy light, leaving UV to pass through. Even if the blacklight consumes 50 watts of power, the light emitted will have much less than 50 watts of power because most of the radiated power will be blocked by the filter.

Different methods of light generation produce different amounts of light given an amount of input electrical power.

UV photons have more energy than IR photons. One way to look at it is that less number of UV photons will be released per second than IR photons (Or intensity of UV light will be less than IR light)

It should be clarified that they do not emit photons of the same frequency, since you specified that one was UV and the other IR, which are quite distinct frequency bands.
otherwise, as the other poster points out, if both bulbs have the same efficiency, which is unlikely using the usual cheap designs, they will emit the same amount of energy. But each photon will not have the same energy: Planck's law says the energy of an IR photon is lower than the energy of each UV photon. So for the energy emitted to be the same, there must be many more IR photons, again, assuming the same efficiency. Lasers could easily be designed to have the same efficiency.

One could also "design" the incandescent bulbs to have the same inefficiency by crippling the more efficient one with a potentiometer or dimmer so it emits the same amount of energy as the worse bulb.

In summary, there is no principled connection between the power and the frequency emitted. But Planck's law does give an exact relation between each individual photon's frequency and its individual energy. So 5o watts in a beam of IR light has more photons per second than a 50 watt beam of UV light.

That said, the crucial point is are we talking about a beam of light or just one big photon? To say "watts" means energy per second, so clearly we are talking about a continuous process. So a more energetic bulb does not have to produce the higher energy by producing higher frequency photons, it could produce more of the same frequency photons. In fact, the frequency of an emitted photon depends on the exact relation between some energy levels in an atom somewhere that emit the photon. So it could be said that if we are talking on the micro-level, and just one event, one photon being emitted, an atom sparking between two widely spaced energy levels will produce more energy because it emits a photon with higher frequency. And conversely, if you wanted to design a one-shot one event emitter, that was only going to emit one photon and then stop, and if you wanted more energy, you would have to go to higher frequencies. But that's not a bulb, and not really watts anymore.