Angular momentum of a circualr polarized EM wave

In an exercise, I am being asked to compute the angular momentum of a circularly polarized wave.

The wave is defined by the four potential: $$\Phi^\mu(x^\nu) = \text{Re} \left\{ \varepsilon^\mu e^{ik_\nu x^\nu} \right\}$$

where $$k_\nu = (k, 0, 0, k)$$ and $$\epsilon_\nu = C(0, 1, i, 0)$$.

I have first computed the field tensor: $$F^{\mu\nu} = \text{Re} \left\{ ie^{ik_\alpha x^\alpha}(\varepsilon^\nu k^\mu - \varepsilon^\mu k^\nu) \right\}$$

and then the stress-energy tensor: $$M^{\mu\nu} = \epsilon_0 C^2 k^\mu k^\nu$$

When I try to apply the definition of the angular momentum tensor $$L^{\mu\nu} = \partial_\sigma(x^\mu M^{\nu\sigma} - x^\nu M^{\mu\sigma})$$

I obtain a completely null tensor. I am not sure about this results, should there be a contribution due to the circular polarization? Am I computing only the orbital angular momentum? How can I derive the spin part?

What you named $$L^{\mu\nu}$$ isn't angular momentum tensor. It's the divergence of density of angular momentum tensor. I'm happy you found zero, as it shows angular momentum is conserved.
If you start from a plane wave, which is infinite in space, you're in trouble, as it looks having an infinite angular momentum (as well as an infinite 4-momentum). The right expression of $$L^{\mu\nu}$$ would be $$L^{\mu\nu}= \int\!d^3\!x\,(x^\mu M^{\nu0} - x^\nu M^{\mu0}).$$ Consider the space components $$L^{ik}$$. From your expression we see that only $$M^{30}\ne0$$ and independent of coordinates. Then $$L^{13} = -L^{31} = \int\!d^3\!x\>x^1 M^{30} \qquad L^{23} = -L^{32} = \int\!d^3\!x\>x^2 M^{30}.$$ You can see that these integrals don't exist. If you try to compute them as limits on an increasing sequence of bounded domains, you can get the result you like better, 0 and $$\infty$$ included.