In general, quantum numbers are labels of irreducible representations of the relevant symmetry group, not primarily eigenvalues of an otherwise simply
defined operator.
But for every label that has a meaningful numerical value in every irreducible representation, one can define a Hermitian operator having it as an eigenvalue, simply by defining it as the sum of the projections to the irreducible subspaces multiplied by the label of this representation. It is not clear whether such an operator has any practical use.
This also holds for the spin. However, one can define the spin in a representation independent way, though not via eigenvalues.
The spin of an irreducible positive energy representation of the Poincare group is $s=(n-1)/2$, where $s$ is the smallest integer such that the representation occurs as part of the Foldy representation in $L^2(R^3,C^n)$
with inner product defined by
$~~~\langle \phi|\psi \rangle:= \displaystyle \int \frac{dp}{\sqrt{p^2+m^2}} \phi(p)^*\psi(p)$.
The Poincare algebra is generated by $p_0,p,J,K$ and acts on this space
as follows (units are such that $c=1$):
spatial momentum: $~~~p$ is multiplication by $p$,
temporal momentum = energy/c: $~~~p_0 := \sqrt{m^2+p^2}$,
angular momentum: $~~~J := q \times p + S$,
boost generator: $~~~K := \frac{1}{2}(p_0 q + q p_0) + \displaystyle\frac{p \times S}{m+p_0}$,
with the position operator $q := i \hbar \partial_p$ and the spin vector $S$ in a unitary irreducible representation of $so(3)$ on
the vector space $C^n$ of complex vectors of length $n$, with the same
commutation relations as the angular momentum vector.
The Poincare algebra is generated by $p_0,p,J,K$ and acts on this space irreducibly if $m>0$ (thus givning the spin $s$ representation), while it is reducible for $m=0$. Indeed, in the massless case, the helicity
$~~~\lambda := \displaystyle\frac{p\cdot S}{p_0}$,
is central in the universal envelope of the Lie algebra, and the possible eigenvalues of the helicity are $s,s-1,...,-s$, where $s=(n-1)/2$. Therefore, the eigenspaces of the helicity operator carry by restriction unitary
representations of the Poincare algebra (of spin $s,s-1,...,0$), which are easily seen to be irreducible.
The Foldy representation also exhibits the massless limit of the massive representations.
Edit: In the massless limit, the formerly irreducible representation becomes reducible. In a gauge theory, the form of the interaction (multiplication by a conserved current) ensures that only the irreducible representation with the highest helicity couples to the other degrees of freedom, so that the lower helicity parts have no influence on the dynamics, are therefore unobservable, and are therefore ignored.
For semisimple groups (and the Poincaré group is not such), the number of
Casimirs (i.e., the number basic generators center of the universal
enveloping algebra) is equal to the dimension its Cartan subalgebra (maximal commuting subalgebra) which is the rank of the algebra. This is called
Chevalley's theorem.
The Poincaré group however is not semisimple (it is a reductive Lie
group given by a semidirect product of a semisimple (Lorentz) and an
Abelian group (translations)), thus this theorem is not valid in this
case (even if both ranks are equal to 2 in the case of the Poincaré
group, there is no general theorem for that).
For such groups the center of the universal enveloping algebra can be
characterized by the Harish-Chandra isomorphism , which is less constructive than the Chevalley's theorem.
There is a way to "understand" why the number of Casimirs of the
Poincaré group is 2. The Poincaré group is a
Wigner-İnönü contraction of the de-Sitter group $SO(4,1)$ which is semisimple and of rank 2.
The Casimirs of the Poincaré group can be obtained from the Casimirs of
$SO(4,1)$ explicitly in the contraction process. This is not a full proof
because group contractions are singular limits, but at least it is a way
to understand the case of the Poincaré group.
Best Answer
The zero component of the pauli Lubanski vector
$W^0 = \epsilon^{0 ijk}J_{ij}p_k = \epsilon^{ijk}J_{ij}p_k $
The angular momentum genrerators
$ j^k = \epsilon^{ijk}J_{ij}$
Thus
$W^0 = j^k p_k = \vec{j}.\vec{p} $
The orbital angular momentum
$ \vec{l} = \vec{x} \times \vec{p}$
is orthogonal to the momentum:
$ \vec{l}.\vec{p} = 0$
And since the total angular momentum is the vector sum of the orbirtal and the spin angular momenta
$ \vec{j} = \vec{l} + \vec{\Sigma}$
Thus
$W^0 = j^k p_k = \vec{j}.\vec{p} = (\vec{j-l}).\vec{p} = \vec{\Sigma}.\vec{p} $
Now, since
$W^0 = \hat{h} p_0$
and for a massless particle
$ p_0 = p$
We obtain:
$ \hat{h} = \frac{\vec{\Sigma}.\vec{p}}{p_0} = \vec{\Sigma}.\hat{p}$
$$\hat{h} = \Sigma.\hat{p}$$
where $\Sigma$ is the spin operator and $\hat{p}$ is the momentum unit vector is a projection along the axis $\hat{p}$ of a spin operator, thus one might expect it to have for a helicity $\lambda$ the eigenvalues $\lambda$, $\lambda-1$, ..., $-\lambda$.
However, the eigenvectors corresponding to all eigenvalues except $\pm \lambda$ are not physical, because they describe longitudinal polarizations which do not exist in free massless particles.
Here is an example of the massless spin-1 case (photon). In this case, we may choose the spin operators as:
$ \Sigma_x = \begin{bmatrix} 0 & 0& 0\\ 0 & 0& -i\\ 0 & i& 0 \end{bmatrix}$
$\Sigma_y = \begin{bmatrix} 0 & 0& i\\ 0 & 0& 0\\ -i & 0& 0 \end{bmatrix}$
$\Sigma_z = \begin{bmatrix} 0 & -i & 0\\ i & 0& 0\\ 0 & 0& 0 \end{bmatrix}$
The action of the Helicity operator on (say), the electric field in the momentum representation is:
$$\hat{h} \vec{E} = i\begin{bmatrix} 0 & -\hat{p}_z & \hat{p}_y\\ \hat{p}_z & 0& -\hat{p}_x\\ -\hat{p}_x & \hat{p}_x& 0 \end{bmatrix}\begin{bmatrix} E_x\\ E_y\\ E_z \end{bmatrix} = i \hat{p}\times \vec{E}$$
Thus:
$$\hat{h}^2 \vec{E} = - \hat{p} \times ( \hat{p}\times \vec{E}) = \vec{E} -\hat{p}(\hat{p}. \vec{E})$$
But, since for a free electromagnetic field:
$$\hat{p}. \vec{E} = 0$$
We get:
$$\hat{h}^2 = 1$$,
and the only admissible eigenvalues are $\pm 1$