If $[\hat A,\hat B]=0$ and they are both non-degenerate, then you every eigenstate of $\hat A$ is an eigenstate of $\hat B$ and vice versa.

If $[\hat A,\hat B]=0$ and $\hat A$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $\hat A$ will be an eigenstate of $\hat B$.

As a simple counterexample to illustrate that last statement, take the operators $$ \hat A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix} \quad\text{and}\quad \hat B = \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}, $$ for which $(1,0,0)^T$ is an eigenstate of $\hat{A}$ but not $\hat B$ even though $\hat A\hat B=\hat B\hat A=\hat B$.

If the information you have is that $[\hat A,\hat B]=0$, $\hat A$ has a degenerate spectrum and $v$ is an eigenstate of $\hat A$ in a space with degenerate eigenvalue, then you cannot make any inferences about its relationship to $\hat B$ $-$ it might be an eigenstate, or it might not.

If $[\hat A,\hat B]=0$ and they are both non-degenerate, then every eigenstate of $\hat A$ is an eigenstate of $\hat B$ and vice versa.

If $[\hat A,\hat B]=0$ and $\hat A$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $\hat A$ will be an eigenstate of $\hat B$.

As a simple counterexample to illustrate that last statement, take the operators $$ \hat A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix} \quad\text{and}\quad \hat B = \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}, $$ for which $(1,0,0)^T$ is an eigenstate of $\hat{A}$ but not $\hat B$ even though $\hat A\hat B=\hat B\hat A=\hat B$.

If the information you have is that $[\hat A,\hat B]=0$, $\hat A$ has a degenerate spectrum and $v$ is an eigenstate of $\hat A$ in a space with degenerate eigenvalue, then you cannot make any inferences about its relationship to $\hat B$ $-$ it might be an eigenstate, or it might not.

If $[\hat A,\hat B]=0$ and they are both non-degenerate, then you every eigenstate of $\hat A$ is an eigenstate of $\hat B$ and vice versa.

If $[\hat A,\hat B]=0$ and $\hat A$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $\hat A$ will be an eigenstate of $\hat B$.

As a simple counterexample to illustrate that last statement, take the operators $$ \hat A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix} \quad\text{and}\quad \hat B = \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}, $$ for which $(1,0,0)^T$ is an eigenstate of $\hat{A}$ but not $\hat B$ even though $\hat A\hat B=\hat B\hat A=\hat B$.

If $[\hat A,\hat B]=0$ and they are both non-degenerate, then you every eigenstate of $\hat A$ is an eigenstate of $\hat B$ and vice versa.

If $[\hat A,\hat B]=0$ and $\hat A$ has a degenerate spectrum, then you are guaranteed the existence of one common eigenbasis. However, you are not guaranteed that every eigenstate of $\hat A$ will be an eigenstate of $\hat B$.

As a simple counterexample to illustrate that last statement, take the operators $$ \hat A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix} \quad\text{and}\quad \hat B = \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}, $$ for which $(1,0,0)^T$ is an eigenstate of $\hat{A}$ but not $\hat B$ even though $\hat A\hat B=\hat B\hat A=\hat B$.

If the information you have is that $[\hat A,\hat B]=0$, $\hat A$ has a degenerate spectrum and $v$ is an eigenstate of $\hat A$ in a space with degenerate eigenvalue, then you cannot make any inferences about its relationship to $\hat B$ $-$ it might be an eigenstate, or it might not.