Showing Two Operators Share an Eigenvector Question 1 - Let $A$ and $B$ be complex $n \times n$ matrices such that $AB = BA^2$, and assume $A$ has no eigenvalues of absolute value 1. Prove that $A$ and $B$ have a common (nonzero) eigenvector. 
Question 2 - Let $A$ and $B$ denote real $n \times n$ symmetric matrices such that $AB = BA$. Prove that $A$ and $B$ have a common eigenvector in $\mathbb{R}^n$. 
For the second question. Suppose $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ so $Av=\lambda v$. Then 
$$A(Bv) = B(Av) = B(\lambda v) = \lambda Bv$$
which shows that the eigenspace corresponding to $\lambda$ is invariant under B. Now what....
 A: For the part that you were missing/asking for Question 2:
Since $B$ is symmetric $B=U^tDU$ for some diagonal matrix $D$ and some orthogonal matrix $U$.
If $E$ is invariant for $B$ then $U(E)$ is invariant for $D$. Therefore $D$ has an eigenvector $u$ there. It follows that $U^tu$ is an eigenvector for $B$ and belongs to $E$. Therefore is is also an eigenvector for $A$.
It is interesting that this part didn't need that $A$ was symmetric. But you used it to ascertain that $A$ does have an eigenvector. Therefore in question 2 one only needs to know that $A$ has an eigenvector.
Question 1:
Assume all eigenvalues of $A$ are zero and $E$ is its kernel. Then $B(E)\subset E$. Therefore there is an eigenvector of $B$ in $E$ and that would be a common eigenvector for $A$ and $B$. It would correspond to the eigenvalues $0$ for $A$ and perhaps to some other eigenvalue for $B$.
Assume now that not all eigenvalues of $A$ are zero. Let $r\neq0$ be one non-zero eigenvalue of $A$ and $E$ its eigenspace.
Then for all $v\in E$, $ABv=BA^2v=r^2Bv$. If $Bv=0$ for some non-zero $v\in E$ we are done, $v$ is eigenvector of $B$ and of $A$. 
Assume then that $Bv\neq0$ for all $0\neq v\in E$. Therefore $r^2$ is eigenvalue of $A$. Now, for $0\neq v\in E$, $AB^2v=BA^2Bv=B^2A^4v=r^4B^2v$. Again, if $B^2v=0$ then $Bv$ is a common eigenvector of $A$ and $B$, of eigenvalue $r^2$ for $A$ and $0$ for $B$. If you assume that $B^2v\neq0$ for all such $v$, then $r^4$ is an eigevalue of $A$.
Continuing in this way we would find either a common eigevector of $A$ and $B$, or a sequence $r,r^2,r^4,...$ of eigenvalues of $A$. Since there are only finitely many eigenvalues we must have $r^m=r^s$ for some $m\neq s$. Since $|r|\neq1$ we must have $r=0$. But we assumed $r\neq0$. Therefore, in this process we should always find a common eigenvalue.
