= Solution
Insert <hydrogen> <energy eigenstates> on either side of the <commutator>. Because $L_z$ is <Hermitian> and has <eigenvalue> $\hbar m$,
$$
0=\langle nlm|[L_z,z]|nl'm'\rangle=\hbar(m-m')\langle nlm|z|nl'm'\rangle.
$$
Thus
$$
\boxed{\langle nlm|z|nl'm'\rangle\ne0\ \Longrightarrow\ m=m'.}
$$
This is a \b[necessary <electric-dipole selection rule>], not a sufficient condition for a nonzero <matrix element>. For example, $\langle200|z|200\rangle=0$ by <parity> even though $m=m'=0$. The angular factor $z=r\cos\theta$ additionally requires $l'=l\pm1$: multiplication of a <spherical harmonic> by $\cos\theta$ gives only harmonics with the same $m$ and adjacent $l$. The <commutator> alone establishes only the <magnetic quantum number> restriction; radial <integrals> and the remaining <electric-dipole selection rules for hydrogen> must still be considered.
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