Can the Klein-Gordon Equation represent Particles with non-zero spin? Every Solution of the Dirac Equation is also a solution of the Klein-Gordon equation.
So the K-G equation does not necessarily represent particles with non-zero spin.
Would it be incorrect to conclude that KG equation relates solely to zero-spin particles?
 A: 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).
A: There is a simple terminological confusion here. What you refer to the "Klein-Gordon equation" is known as the "wave equation" or "d'Alembert equation" with a mass term. Fields of any spin may be the variables in such an equation. Of course it's useful to realize the similarities between the equations governing the scalar field and equations governing other fields. The essence is always the same.
However, the Klein-Gordon equation is, by definition, the wave equation for a spinless, scalar field. In particular, the Klein-Gordon field has to be a scalar field, by definition. So the answer to your question is No. If the equation contains a field with a non-zero spin, then it is not the Klein-Gordon equation. This claim is no different from the claim that the formula $s=vt$ for the distance as a product of the velocity and the time isn't called the Ohm's law. Ohm's law has the same mathematical form – something is the product of other two things, $U=RI$ – but it actually requires the objects that enter the product to have some particular properties. That's why $s=vt$ isn't Ohm's law.
A: In addition to the answers above, I would like to add what goes wrong if you try to use Klein Gordon equation for higher spin. In this case, you always get ghost particles, you get particles whose contribution to the probability is negative.
This issue is best explained by trying this trick with a vector. Consider a four-component vector, with each component separately obeying the KG equation. In this case, you get
$$ S = \int  \partial_\mu A_\nu \partial_\mu A_\nu - {m^2\over 2} A_\mu A\mu d^4x$$
and the kinetic term on the time component of A is opposite sign to all the space components. This means you have a timelike ghost field, or three spacelike ghost fields, and the equation is not describing a reasonable free field theory.
Nevertheless, this thing (in the massless case) is what you get in QED when you impose Feynman gauge (or in t'Hooft gauge for the nonabelian theory). The ghost time component doesn't contribute to physical processes by gauge invariance, and this means that the negative propagator states are not propagating degrees of freedom.
This disease affects the KG equation with any spin other than 0, so the other wave equations are the proper ghost-free trucations of the Klein Gordon equation, which is best viewed as only part of the equation of motion--- the mass shell projection.
A: Each free particle or field (each component of corresponding field or wave-function) must satisfy the Klein-Gordon equation, because it corresponds to relativistic energy-momentum relation (or, more formally, refers to the Casimir operator $p^{\mu}p_{\mu}$ of the Poincare group). 
But each free particle or field must satisfy relation $W^{\mu}W_{\mu} \Psi = -ms(s + 1)\Psi$, where $W_{\mu}W^{\mu}$ is the second Casimir operator (refers to the spin degrees of freedom). There is $2s + 1$ spin degrees of freedom (energy is equal for corresponding states). So you need to add some relations to your function (it may refer to the spinor $\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{m}}$ or, in particular cases, to 4-tensor $A_{\mu_{1}...\mu_{n}}$), which left only $2s + 1$ independent components (or only one for massless particles). For the free particles almost all other charges lose significance, because they are important only in an interaction cases. So we usually say that free particle has mass $m$ and spin (helicity) $s$.
For example, let's have spin-s massive particle. This case corresponds to s-rank 4-tensor $A_{\mu_{1}...\mu_{s}}$ with Klein-Gordon equation for each component
$$
\tag 1 (\partial^{2} + m^{2})A_{\mu_{1}...\mu_{s}} = 0
$$ 
and conditions
$$
\tag 2 \partial_{\mu_{k}}A^{\mu_{1}...\mu_{k}...\mu_{s}} = 0,\quad k = 1, ..., s,
$$ 
$$
\tag 3 A^{\mu_{1}...\mu_{k}...\mu_{l}...\mu_{s}} = A^{\mu_{1}...\mu_{l}...\mu_{k}...\mu_{s}}, \quad k = 1, ..., s,
$$
$$
\tag 4 A_{\mu_{k}}^{\quad \mu_{1}...\mu_{k}...\mu_{s}} = 0, \quad k = 1, ..., s.
$$
