The vacuum Dirac equation automatically implies the Klein-Gordon equation. It means that every solution to the vacuum Dirac equation is automatically a solution to the Klein-Gordon equation.
The converse of course doesn't hold. The most basic reason is that the Klein-Gordon equation should really act on scalars, a single bosonic field, while the minimum number of components for the $d=4$ Dirac equation is four (and they should be fermionic fields). So a general (or generic) valid solution to the Klein-Gordon equation is a valid solution to the Klein-Gordon equation (this much is a tautology, but you were asking about it), but it is not a solution to the Dirac equation.
Even if you combine 4 solutions to the Klein-Gordon equation, declare that they are 4 components of a Dirac spinor, and ask whether they solve the Dirac equation, the answer is No. It's because the Dirac equation is really "stronger" than the Klein-Gordon equations for its components. Effectively, the Dirac equation is first-order while the Klein-Gordon equation is second-order. The Dirac equation implies certain correlations between the spin (up/down) of the particle and the sign of the energy (positive/negative). The quadruplet of Klein-Gordon equations allows all combinations of spin up/down and the sign of the energy.
However, the most general quadruplet of solutions to the Klein-Gordon equation may be written as a solution of the Dirac equation with a positive mass and a solution to the Dirac equation with a negative (opposite) mass.
The Dirac equation describes spin-1/2 (and therefore "fermionic") particles such as electrons, other leptons, and quarks, while the Klein-Gordon equation describes spin-0 "scalar" (and bosonic) particles such as the Higgs boson. However, before they do the proper job, the "wave functions" have to be promoted to full fields and these fields have to be quantized.
Spin is a property of the representation of the rotation group $SO(3)$ that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation.
The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also accounts for discrete symmetries).
The components of every free field satistfy the Klein-Gordon equation, irrespective of their spin. In particular, every component of the Dirac equations solves the Klein-Gordon equation.
Indeed, the Klein-Gordon equation only expresses the mass shell constraint and nothing else. Spin comes in when one looks at what happens to the components.
A rotation (and more generally a Lorentz transformation) mixes the components of the Dirac field (or any other field not composed of spin 0 fields only), while on a $k$-component spin 0 field, it will transform each component separately.
In general, a Lorentz transformation given as a $4\times 4$ matrix $\Lambda$ changes a $k$-component field $F(x)$ into $F_\Lambda(\Lambda x)$, where $F_\Lambda=D(\Lambda)F$ with a $k\times k$ matrix $D(\Lambda)$ that depends on the representation. The components are spin 0 fields if and only if $D(\Lambda)$ is always the identity.
Best Answer
KG equation written in the momentum space is just a mass-shell condition, so it holds for any particles that are on-shell (e.g. gauge-fixed Maxwell equations in the vacuum are $\square A_{\mu}=0$, which is just KG equation for the massless field).