Bound vs. extended states
In condensed matter physics bound states are the states with the wave functions decaying towards infinity, as opposed to the extended states, as, e.g., plane waves. In this sense the definitive answer to the question can be given only by the exact diagonalization of the Hamiltonian and studying the behavior of the wave functions. Without that the question is necessarily vague, and the discussions of the bound states usually carry qualitative character.
Gaps between the states
Another way to define bound states is as isolated states, i.e., they are not a part of a continuous spectrum. In non-interacting case (or in the spectrum of a diagonalized Hamiltonian), these are immediately seen from the dependence of $\epsilon_k$ on $k$. In other words, they are separated by a gap from the adjacent bound states or the continuum. In this picture, including the interactions broadens the states, and whether they remain (quasi-)bound depends on how the broadening compares to the gap. When the broadening is very strong we say that the bound states wash out, becoming continuous.
Gap opening
Finally, there is a class of problems where we are interested in appearance of bound states. In some cases one can re-express the problem in terms of an effective Schrödinger equation (e.g., by summing the ladder diagrams - see Fetter&Walecka for a clear presentation). In this case one distinguishes bound and extended states by the propreties of the solutions of thsi effective Schrödinger equation (i.e., decaying to infinity or not).
Another case is gap opening, which can be studied by a number of techniques, e.f., renormalization group.
Clarification about the spectral function
The textbooks on QFT in condensed matter physics teach us that the partcle (and multiparticle) excitations appear as singularities in the Green's function. It is necessary to stress here that Green's function is not the same thing as the spectral function (like the one given in the question). In fact, the spectral function is analytical, i.e., it does not have singularities. It is convenient to use mathematically, but it is less suitable for the question that interests us. In fact, although it gives a good intuition when discussing broadening of states or their delocalization, as I discussed above, it can be very misleading in terms of finding the true excitations, since a lot of $\omega$-dependence, and all the new physics due to the interaction, is hidden in the self-energy.
If we work with the Green's function, then its singularities will be either isolated poles or branch cuts - corresponding respectively to isolated (likely bound) states and the continuous spectrum.
Another point - the spectral function given in the question is that for a one-particle Green's function. If we are interested in two-particle bound states, operating with a two-particle Green's function and the corresponding spectral representation.