Are for example all eigenstates of $L_z$ also eigenstates of $L^2$?
No. There are plenty of situations where you can have eigenstates of $L_z$ that are not eigenstates of $L^2$.
The simplest example is probably the one you have in mind and the first one you were introduced to (if your route into quantum mechanical angular momentum followed the normal steps), namely, the hydrogen atom. (Or, by extension, any particle in 3D in a central potential.)
Here, the eigenstates are normally denoted as $|n,\ell,m⟩$, where $n$ is the principal quantum number, $\ell$ labels the eigenvalue of $L^2$ and $m$ labels the eigenvalue of $L_z$.
The important thing to note here is that for any given value of $m$, there are multiple different $\ell$ subspaces that contain states with that eigenvalue of $L_z$. This means that you can form superpositions with the chosen $m$ but with different $\ell$: say, something like
$$
|3,1,1⟩+|3,2,1⟩ = |3p_1⟩ + |3d_1⟩,
$$
which has a well defined $L_z=1$, but which does not have a well-defined $L^2$.
Now, as pointed out in the existing answer on this thread, once you get more into the details of quantum mechanics and things get correspondingly more abstract (and, particularly, once you learn to contextualize the quantum mechanical theory of angular momentum as one application of group representation theory), then it becomes more common to consider Hilbert spaces that contain a single eigenvalue of $L^2$ (in technical language, "irreducible representations", often abbreviated as "irrep"s).
If you are within that setting, then all eigenstates of $L_z$ are also eigenstates of $L^2$, simply because all the states in the space under consideration are eigenstates of $L^2$, and they all have the same eigenvalue. However, from the way you phrased the question, I think it's unlikely that you're thinking about a setting like this one.
Regarding the second half of your question,
If this is the case, does this mean that $L^2$ has more eigenstates than $L_z$ since it also shares eigenstates with $L_x$ and $L_y$? Do the all eigenstates of $L^2$ consist of the union of the eigenstates of $L_x$, $L_y$ and $L_z$?
there is largely no objective way of comparing which operator has "more" eigenstates. In any nontrivial situation (i.e. when you have more than one eigenspace of $L^2$ in your Hilbert space), there is an infinity of eigenstates of $L^2$ that are not eigenstates of $L_z$ (and not eigenstates of $L_x$ or $L_y$ either), and there is also an infinity of eigenstates of $L_z$ which are not eigenstates of $L^2$.
There are two trivial cases in this regard,
- the case of $\ell=0$, where there is a single state involved; and
- the case of $j=1/2$, where every eigenstate of $J^2$ is an eigenstate of a component $\hat{\mathbf n}\cdot\mathbf J$ of $\mathbf J$ along some direction $\hat{\mathbf n}$ (but not necessarily the restricted set of Cartesian coordinates along some pre-chosen set of axes);
but for anything bigger than that, the sets in question are just different enough that you cannot really compare them.