Perhaps this will be illuminating. Consider the operator $\vec n\cdot \vec L$ corresponding to measuring angular momentum in an arbitrary direction specified by a unit vector $\vec n$, then I make the following:
Claim. For any two unit vectors $\vec n_1$ and $\vec n_2$, the spectra (sets of eigenvalues) of $\vec n_1\cdot\vec L$ and $\vec n_2\cdot \vec L$ are the same.
I'll prove this claim below, but before I do, let's think about what this means. If we were to measure the component of angular momentum pointing in any direction we choose, this result says that we always get the same, discrete sets of possible measured values. In particular, if we choose to call a certain direction in space our $z$-axis, or if we rotate our coordinate system and call another direction in space our $z$-axis, then the set of possible measured values we will get for the $z$-component of the angular momentum will be the same in both cases.
Proof of claim.
Since $\vec n_1$ and $\vec n_2$ are unit vectors, there is a rotation (special orthogonal transformation) $R$ for which
$$
\vec n_1 = R\vec n_2
$$
It follows that
\begin{align}
\vec n_2\cdot \vec L
&= (R^t\vec n_1)\cdot \vec L \\
&= \vec n_1\cdot (R\vec L) \\
&= (n_1)_iR^i_jL^j \tag{$\star$}
\end{align}
Now, recall that $\vec L$ is a vector operator; this means that there exists a unitary representation $U$ of the rotation group acting on the Hilbert space such that
$$
U(R)^{-1} L^iU(R) = R^i_jL^j
$$
so that for any unit vector $\vec n$, one has
$$
U(R)^{-1} n_iL^iU(R) = n_iR^i_jL^j
$$
which in vector notation can be written as
$$
U(R)^{-1} (\vec n\cdot\vec L) U(R) = \vec n\cdot(R\vec L) \qquad \tag{$\star\star$}
$$
Upon combining $(\star)$ and $(\star\star)$, we find that
$$
\vec n_2\cdot\vec L = U(r)^{-1} (\vec n_1\cdot\vec L) U(R)
$$
This shows that the two operators $\vec n_1\cdot\vec L$ and $\vec n_2\cdot\vec L$ are related by a unitary similarity transformation, so they have the same spectra.
