= Solution
The total <angular momentum operator> is
$$
\mathbf L^2=L_x^2+L_y^2+L_z^2.
$$
Using the cyclic <orbital angular momentum commutation relations>,
$$
[L_z,L_x^2]=i\hbar(L_yL_x+L_xL_y),
\qquad
[L_z,L_y^2]=-i\hbar(L_xL_y+L_yL_x),
$$
while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,\mathbf L^2]=0}$. In Cartesian coordinates,
$$
\boxed{L_z=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)}.
$$
Because $L_z(x+iy)=\hbar(x+iy)$ and it annihilates $z$ and every radial function,
$$
L_z[(x+iy)^mz^nf(r)]=m\hbar(x+iy)^mz^nf(r).
$$
Replacing $i$ by $-i$ gives eigenvalue $-m\hbar$.
The six-dimensional space splits into the five trace-free quadratic <spherical harmonics> with $\ell=2$ and one radial state with $\ell=0$. A simultaneous eigenbasis for $L_z$ and $\mathbf L^2$ is
$$
\begin{array}{c|c|c}
\text{state}&L_z&\mathbf L^2\\ \hline
(x+iy)^2f(r)&2\hbar&6\hbar^2\\
(x+iy)zf(r)&\hbar&6\hbar^2\\
(2z^2-x^2-y^2)f(r)&0&6\hbar^2\\
(x-iy)zf(r)&-\hbar&6\hbar^2\\
(x-iy)^2f(r)&-2\hbar&6\hbar^2\\
(r^2-3)f(r)&0&0
\end{array}
$$
where the eigenvalue $\ell(\ell+1)\hbar^2$ of $\mathbf L^2$ was used. Every displayed state is a complex <linear combination> of the original six energy eigenstates, so it is still an energy eigenstate at the same level.
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