This is a rather complicated question in the sense that a proper answer needs to use nuclear structure physics terms.
Trying to be concise at the same time, it is such that the alpha decays from one superheavy atomic nucleus to its daughter is "simplest" if the number of both protons, Z, and neutrons, N, is even - say 292Lv (116 protons and 176 neutrons).
For such nuclei, it is by far most likely that the alpha-decay occurs between the "ground states" of these two nuclei. This means that the alpha particle is almost always emitted wit the same energy. Thus, the decay energy of the alpha particle can to some extent be considered a fingerprint of such a decay. This argument has, at least in part, being used when the discoveries of elements 114 and 116 were approved.
The alpha decay of odd-Z (or odd-N or both odd Z and N) is most often more complicated. This is because it possibly involves several excited nuclear states in both mother and daughter nucleus. Therefore, the alpha-decay fingerprint is typically (if not definitely) lost as the alpha-decay energy is not always the same.
In turn, the excited states in nuclei decay by electromagnetic radiation - gamma rays (one kind of photons) - or by transferring their excitation energy to the surrounding electrons of the superheavy atom. In the latter case the atom gets excited while the nucleus de-excites. Following upon this, the atoms de-excite by sending out X-rays (also a kind of photon, but from the atom). So either way, gamma-ray decay of the nucleus or X-ray decay of the atom, one needs to measure these photons to be able to "fingerprint" again. On top, energies of X rays are characteristic for an element (H.G.J. Moseley's law from 1913).
As ever so often in science, complications (in nuclear structure) give rise to opportunities (X-ray fingerprinting).