Part of your question is how to define bound states in QFT. Just as in QM, a bound state in QFT is a pole of the nonperturbative T-matrix (, i.e., the S-matrix minus 1 stripped from a delta function), and its possible wave functionals are the functionals in the range of the projector linear operator obtained as the residual of the T-matrix at the pole. (A proof is given at the end of this post. For related information see my entry ''Bound states in relativistic quantum field theory'' of Chapter B6: Bound states and applications of my theoretical physics FAQ.)

In particular, elementary particles as conventionally modelled by quantum fields as their elementary collective excitations are bound states iff they can appear as asymptotic states of the theory. Thus electrons and photons are bound states of QED or the standard model, but (because of confinement) quarks in QCD or the standard model are not.

Pions exist as asymptotic states of QCD, hence form a bound state. Since the contribution of the singe-particle fields to their wave functionals vanishes, they are nonelementary (composite) collective excitations. The same holds for all mesons and baryons. In general, an asymptotic $k$-particle composite makes its first appearance in the $k$-particle sector of the T-matrix, i.e., in S-matrix elements with $2k$ external legs.

Positronium is not a bound state of QED but only a resonance (unstable state), as it is represented by a complex pole of the analytic continuation of the T-matrix into the second (unphysical) sheet. The same holds for the pion when regarded as a state of the standard model, since pions decay under the weak force.

In terms of Ron Maimon's parable, made slightly more precise, **all particles** (whether elementary or composite) are to quantum fields like sound is to the pressure field of air.

Proof of the T-matrix property used above: The Lippman-Schwinger equation for the T-matrix gives $V=(1-VG_0)T=((E-H_0)G_0-VG_0)T=(E-H)G_0T$, hence $G_0T=(E-H)^{-1}V$. Now the spectral theorem implies that if $\psi$ is an eigenvector for a simple eigenvalue $E_0$ of $H$ then the spectral decomposition of $(E-H)^{-1}$ contains a term proportional to $(E-E_0)^{-1}\psi\psi^*$, hence $G_0T$ contains a term proportional to $(E-E_0)^{-1}\psi\psi^*V$. But this means that the residue of $E_0$ is $\psi\psi^*V$. The proof in the degenerate case is similar, but gives a higher rank residue, which has a higher-dimensional eigenspace as range.