Let us consider an electron in an infinite square well. As we know that the electron has a spin=$1/2$ . The spin operator ($z$-direction) has two eigenvectors which span the vector space. But if we solve for the eigenvectors of Hamiltonian, We get infinite basis vectors which span the space. But in linear algebra, we cannot have two basis sets with different number of elements.
The spin operator $S_z$ has two eigenvalues, and its eigenvectors span the whole state space, but that doesn't mean it has two eigenvectors.
In your case, the full state space is spanned by states of e.g. definite spin and position, or spin and momentum, or if you want something like spin, energy and sign of the momentum, etc. Since all these values can be assumed independently (i.e. all combinations give a valid and different state), the full state space is the tensor product of the individual state spaces, in your example the abstract two dimensional spin state space, and the infinite dimensional position space.
That means that, as you asked in your comment, indeed a particle that has a definite spin can be in any superposition of position eigenstates.