Suppose two operators $M$ and $N$ commute, $[M,N]=0$. Then if $M$ has an eigenvector $\vert v\rangle$ with non-degenerate eigenvalue $\lambda_v$, we will show that $\vert v\rangle$ is also an eigenvector of $N$. \begin{eqnarray*} M\vert v\rangle &=& \lambda_v\vert v\rangle\\ NM\vert v\rangle &=& MN\vert v\rangle=\lambda_vN\vert v\rangle\\ \end{eqnarray*} The last equality shows that $N\vert v\rangle$ is also an eigenvector of $M$ with the same non-degenerate eigenvalue $\lambda_v$. But if this is true, then $N\vert v\rangle$ must be proportional to $\vert v\rangle$, i.e. \begin{equation} N\vert v\rangle = \alpha \vert v\rangle \end{equation} which is just the statement that $\vert v\rangle$ is also an eigenvector of $N$ with eigenvalue $\alpha$.

Note that if $\lambda_v$ were a degenerate eigenvalue, then we would not have been able to assume that $N\vert v\rangle$ is proportional to $\vert v\rangle$.