One tool for finding your way towards a proof is to construct some representation matrices for a few different $L_{x,y,z}$ and see how you feel about them. Here’s a representation for spin two, which gives you a sense of how it goes. The size of the matrix is $2\ell+1$. The $z$-axis matrix has the eigenvalues on the diagonal already,
$$
L_z = \left(\begin{array}{ccccc}
\ell
\\ &\ell-1
\\ && \ddots
\\ &&& -\ell+1
\\ &&&& - \ell
\end{array}\right),
$$
and the raising and lowering matrices are one-off-the-diagonal,
$$
L_+ = \left(\begin{array}{ccccc}
0& \sqrt{2\ell} &
\\ &0& \sqrt{\cdots} &
\\ &&\ddots
\\ && \cdots & 0& \sqrt{2\ell}
\\ &&&&0
\end{array}\right)
\qquad
L_- = (L_+)^T
\renewcommand{\ket}[1]{\left|{#1}\right>}
$$
so that $L_\pm\ket{\ell,m} = \sqrt{\ell(\ell+1) - m(m\pm1)} \ket{\ell, m\pm1}$.
From these raising and lowering matrices $L_\pm$ you construct
\begin{align}
2 L_x &= L_+ + L_-
\\
2i L_y &= L_+ - L_-
\end{align}
and then just find the eigenvectors. For spin-half (which isn’t allowed for orbital angular momentum $L$, but the algebra is similar), the $x$-axis eigenvectors are ${1}\choose{\pm1}$, with eigenvalues $\pm1$, and the $y$-axis eigenvectors are $1\choose{\pm i}$, with eigenvalues $\pm i$. Those basis vectors are orthogonal to each other. However, it is the case that the eigenstates of $\sigma_x$ have expectation value $\left<\sigma_y\right> = \left<\sigma_z\right> = 0$.