There are no elementary Goldstone bosons in nature. That's why they are not given in tables of elementary constituents which are leptons, quarks, gauge bosons, and the Higgs boson in the Standard Model. It's likely that this is however not the complete zoo as there's a lot of evidence that there is "dark matter" not yet identified as measured particles but only seen by its gravitational effects on stars in galaxies and as part of the cosmological standard model to fit the high-precision data on the cosmic microwave background.
A Goldstone boson always occurs if you have spontaneous symmetry breaking of a global symmetry, which implies that the ground state of the theory is degenerate and does not obey the symmetry that governs the Hamiltonian and thus the dynamics of the system.
One example realized in nature is only approximate but very important: The strong interaction on the fundamental level is described by Quantum Chromodynamics, which however can be treated with perturbative methods only for collisions at very high energies, where the "running coupling" is small (aysmptotic freedom). In the low-energy region, you cannot even observe the fundamental constituents, the quarks and gluons due to confinement, i.e., the asymptotic free states (which we see as particles) are color-charge neutral bound states, the socalled hadrons. Most of them are either mesons (a quark-antiquark boundstate) and baryons (three-quark bound states).
Fortunately in the light-quark sector (up and down quarks) QCD has an approximate chiral symmetry, because the quark masses of a few MeV are smaller than the typical hadronic masses like the proton mass, so that you can approximately treat these quarks as massless. In this limit QCD has a chiral ##\mathrm{SU}(2)_{\text{L}} \times \mathrm{SU}(2)_{\text{R}}## symmetry, which would imply (if realized in the usual "Wigner-Weyl mode") that for each hadron there should be another hadron of the same mass but opposite parity. That's however not observed in nature. The reason is that the strong interaction is so strong that the vacuum (i.e., the ground state of QCD) is not the naive "perturbative" vacuum state but a state, where a socalled quark condensate forms, i.e., ##\langle \bar{\psi} \psi \rangle \neq 0##, where ##\psi## are the quark fields. Now this quark condensate is not invariant under chiral transformations but only under the socalled isovector transformations. This means the chiral symmetry is broken to ##\mathrm{SU}(2)_{\text{V}}##. Now QFT implies that then there must as many massless Goldstone bosons (or better Nambu-Goldstone bosons) as the dimension of the symmetry groups. The chiral symmetry is a 6-dimensional symmetry group, the subgroup which is a symmetry of the ground state is 3 dimensional, which implies that there are 3 massless modes, and these are in our context identified with the pions.
Now the pions are not really massless but have a mass of about 140 MeV. That's due to the fact that the u- and d-quarks are not really massless. So in addition to the above described spontaneous breaking of chiral symmetry it's also slightly explicitly broken, but this you can consider as a correction you can imply by perturbation theory. So the pion acquires its mass due to this explicit breaking. In this sense it's approximately the Goldstone bosons of the approximate chiral symmetry of the light-quark sector of QCD. This is very important, because it tells us how to build effective quantum field theories for hadrons, i.e., the low-energy sector of QCD with the observable hadronic degrees of freedom described by quantum fields.