Why is m_j not a good quantum number in strong-field Zeeman effect?

[Happiness]

TL;DR Summary

When solving for the correction to the Hamiltonian due to strong-field Zeeman effect (using perturbation theory), why is m_j not a "good" quantum number, given that J_z is conserved too?

This textbook claims $m_j$ is not a "good" quantum number because the total angular momentum (of an electron of a hydrogen atom placed in a strong uniform magnetic field) is not conserved. I don't understand why $m_j$ is not a "good" quantum number.

Since $J=L+S$, $J_z=L_z+S_z$.
Since $L_z$ and $S_z$ are both conserved, so is $J_z$.
$J_z$ commutes with $H'_Z$ too.
So shouldn't $m_j$ be a "good" quantum number too?

The phrase "good quantum number" relates to the following theorem in perturbation theory:

The book is "Introduction to Quantum Mechanics", 2nd edition, by David Griffiths.

 

[PeterDonis]

Since $L_z$ and $S_z$ are both conserved, so is $J_z$.

But $J$ is not, and $m_j$ is a quantum number for $J$, not $J_z$.

 

[Happiness]

But $J$ is not, and $m_j$ is a quantum number for $J$, not $J_z$.

This is the remaining part of the section in the book:

From the sentence below [6.81], we can see that eigenstates of $S_z$ and $L_z$ were used as the "good" states $\ket{nlm_lm_s}$ in the perturbation theory in [6.80].

So my question is, aren't eigenstates of $J_z$ "good" states too?

The book did not define quantum numbers explicitly. From what I understand from the book, since $m_j$ is the eigenvalue of operator $J_z$, ie, $J_z\psi=\hbar m_j\psi$ (where $\psi$ is an eigenstate of $J_z$), then $m_j$ is the quantum number for $J_z$. This is how I understand it. ($m_j$ is the eigenvalue apart from a factor of $\hbar$.)