Adapting Cosmas Zachos' comment into an answer.
We say that a field is "free" if the Lagrangian is quadratic in the field and its derivative, eg. for a field $\phi^a$ with some kind of index $a$, the Lagrangian looks like $$ \mathcal L=Q_{ab}^{\mu\nu}\partial_\mu\phi^a\partial_\nu\phi^b+R_{ab}\phi^a\phi^b, $$ where $Q$ and $R$ are field-independent coefficients (in special relativity, usually constant, but in GR it isn't).
The first term is called the kinetic term, the second term is the mass term. It is named as such because upon quantization of the free field, the field quanta will obey a relativistic dispersion relation $E^2=p^2c^2+m^2c^4$ where the mass $m$ is related to the $R_{ab}$ coefficient.
It is usually assumed that the Lagrangian for interacting fields is polynomial in the fields (but if not then I guess one still usually looks at a power series expansion), and further terms containing fields are interaction terms. Eg. $$\mathcal L=Q^{\mu\nu}_{ab}\partial_\mu\phi^a\partial_\nu\phi^b+R_{ab}\phi^a\phi^b+S_{abcd}\phi^a\phi^b\phi^c\phi^d$$ is a Lagrangian where the field $\phi$ has a quartic self-interaction, which is represented by the term involving $S$.
As far as I am aware, if a Lagrangian contains more than 2 instances of derivatives then it is anathema, so these interaction terms are usually field-dependent but not field derivative dependent.
A (vector-) gauge boson is mathematically a connection on a principal fibre bundle, so locally it is a Lie algebra valued $1$-form $A=A^a_\mu T_a\otimes\mathrm dx^\mu$, where $T_a$ are a set of generators for a Lie algebra.
On two overlapping trivialization patches, the local gauge fields transform as $$ A^\prime=\mathrm{Ad}_{g^{-1}}A+g^\ast\Xi, $$ where $g$ is a Lie group ($G$) valued function on the overlap, and $\Xi$ is the left-invariant Maurer-Cartan form on $G$. For matrix groups we can also write $$ A^\prime=g^{-1}Ag+g^{-1}\mathrm dg. $$
The expression $$ F=\mathrm dA+\frac{1}{2}[A\wedge A],\ F=F^a_{\mu\nu} T_a\otimes\mathrm dx^\mu\wedge\mathrm dx^\nu \\ F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A_\mu^a+C^a_{bc}A^b_\mu A^c_\nu $$is the local form of the curvature of the connection, it transforms homogenously as $F^\prime=\mathrm{Ad}_{g^{-1}}F$.
We want Lagrangians to be gauge-invariant, and (aside from some special cases like Chern-Simons terms, which is only weakly gauge invariant) a Lagrangian of the form $$ \mathcal L=\kappa F^a_{\mu\nu}F_a^{\mu\nu} $$ is the only gauge invariant option. Here the greek indices are raised and lowered by the metric tensor, the latin Lie algebra indices are raised and lowered by the Cartan-Killing form or some other Ad-invariant inner product, and $\kappa$ is an irrelevant constant.
If we expand the Lagrangian, then we get schematically (!) $$\mathcal L \sim (\partial A)^2+C\partial A A^2+C^2A^4.$$ The first schematic term is the kinetic term, the second and third are interaction terms (the second term involves only two powers of $A$ but also a $\partial A$ so it is not a mass term).
From this we can deduce some things.
Masslessness: The gauge invariant Lagrangian has no mass term, so the gauge boson is massless. Any attempts to artificially add a mass term of the form say $\sim m^2A_\mu^a A_a^\mu$ ruins gauge invariance.
Charge: Without constructing/defining "charge" explicitly, we see that the Lagrangian for the gauge field has two interaction terms, both involving the structure constants $C^a_{bc}$. This means that a general gauge boson interacts with itself, so we say it has charge, because charged particles are the ones who "feel" the interaction (here charge is not necessarily electic, but the charge of the given interaction, so to speak). However, if $C^a_{bc}=0$, which means that the Lie group is Abelian (all elements commute), then both interaction terms vanish. So an Abelian gauge field doesn't self-interact, so we say that its quanta is not charged.
Spin: This was also mentioned by OP, but is rather tangentially related. In a given Lorentz frame, the spatial components of $A^a_{\mu}$ (eg. $A^a_x,A^a_y,A^a_z$) transform as a 3-vector under spatial rotations, which is a spin-1 degree of freedom. The time component $A^a_0$ doesn't transform under spatial rotations, so it is a spin-0 degree of freedom. Gauge invariance however makes the spin-0 degree of freedom pure gauge, so the field quanta are particles of spin 1.
