Question about the non-renormalization of conserved currents

Question

It is asserted in many places, e.g, sec. 12.4 in Peskin&Schroeder, that the current $J^\mu=\bar\psi\gamma^\mu\psi$ needs no renormalization.

The following diagram

contributes a divergence of $Z_1$-like, which is equal to a $Z_2$-like divergence by Ward identity. Then this divengence can be removed by the renormalization factor $\sqrt{Z_2}$ of $\psi$ in $J^\mu$. So there's no need for a further renormalization of $J^\mu$ as asserted.

However, the following diagram

seems to contribute a $Z_3$-like divergence. $J^\mu$ has to mix with $A^\mu$ to remove this divergence, contradicting that $J^\mu$ needs no renormalization. Is there any problems?

On the other hand, the equation of motion reads $$ Z_3[\partial^2 A^\mu - \partial^\mu(\partial\cdot A)] = - Z_1 e J^\mu \,.$$ While $A^\mu$ is already renormalized, $J^\mu$ also needs no renormalization as asserted. So we should have $Z_3=Z_1$ up to finite terms, which is not correct?

Answer 1

It seems that the current does mix with $A^\mu$, or $\partial_\nu F^{\nu\mu}$ more precisely, as discussed in this paper written by J. C. Collins et al. (I fount it from another question in PSE). I quote from the abstract:

It is commonly asserted that the electromagnetic current is conserved and therefore is not renormalized. Within QED we show (a) that this statement is false

and

The current mixes with the four-divergence of the electromagnetic field-strength tensor.