A pseudoscalar Goldstone boson, $\pi(x)$, is protected by a shift symmetry: it shows up with a derivative in its interaction terms in a Lagrangian. As a pseudoscalar, we may also write it with the usual $i\gamma^5$ interaction. There are thus two ways to encode the interaction:
Shift symmetry manifest: $$\mathcal L = \left(\frac{\partial_\mu \pi}{v}\right)\bar\Psi\gamma^\mu \gamma^5\Psi$$
Pseudoscalar manifest: $$\mathcal L = g \pi \bar\Psi i\gamma^5 \Psi$$
These two are related by the equation of motion, so that $g = 2qm/f$, where $m$ is the fermion mass, $f$ is the order parameter of symmetry breaking, and $q$ is the charge with respect to the broken axial symmetry.
My question is: In the shift-symmetry manifest form, we know that fermion loops do not generate a pseudoscalar mass. However, pseudoscalar manifest form of the interaction looks like a generic pseudoscalar interaction with no symmetry protecting $\pi$ from receiving mass corrections from $\Psi$ loops. So:
How do we see that $\pi$ is protected by a symmetry when we write the interaction in the manifestly pseudoscalar format?
Conversely, I could write any pseudoscalar interaction as $ g \pi \bar\Psi i\gamma^5 \Psi$ Does this mean that I can use the fermion equation of motion to convert any pseudoscalar interaction into one that has a shift symmetry?
Details follow, but the main question is stated above.
Example
Set up: Goldstone interaction with fermions
We show how to convert between the shift-symmetric and pseudoscalar forms of the interaction. For simplicity, assume the case of a global, internal, compact U(1) symmetry that is spontaneously broken by a field $H$ that obtains a vev $\langle H \rangle = f$. Let the theory contain a left-handed fermion $\psi_L$ and a right-handed fermion $\psi_R$. Assume axial U(1) charges such that
$$Q[H] = 2q\\ Q[\psi_L] = q\\ Q[\psi_R] = -q$$
Then we may write out the theory with a Yukawa interaction:
$$\mathcal L_\text{Yuk} = y H^*\bar\psi_L \psi_R + \text{h.c.}$$
We now "pull out the Goldstone fields" from the fields. In order to do this, we transform each field $\Phi \in \{H,\psi_L,\psi_R\}$ by the spontaneously broken symmetry:
$$ \Phi = e^{iq_\Phi \epsilon} \Phi'$$
On the right-hand side, $\Phi'$ is understood to be the field with no Goldstone component. The Goldstone lives in the exponential. For the U(1) case, $\epsilon$ is the transformation parameter, and $q_\Phi$ is the U(1) charge of the $\Phi$.
Then we simply promote the transformation parameter to the Goldstone field, $\epsilon \to \pi(x)/f$. This is a nonlinear transformation to help identify the Goldstone interaction (Sec 19.6 of Weinberg Vol II, or CCWZ II). This gives
$$ \Phi(x) = \exp\left(iq_\Phi \frac{\pi(x)}{f}\right) \Phi'$$
When we do this, $$\mathcal L_\text{int} = y H'^* e^{-2iq} \bar \psi_L' e^{iq} e^{iq} \psi_R' + \text{h.c.} = y H'^*\bar \psi_L' \psi_R' + \text{h.c.} $$
The Goldstone has been completely removed from the Yukawa term and doesn't show up there. This is a consequence of U(1) conservation of the Lagrangian term. Where did the interaction go? We know that the Goldstone must have a derivative interaction, so the natural place to look is the fermion kinetic term.
Writing the kinetic terms with implicit projection operators (alternatively, you can replace $\gamma^\mu$ with $\sigma^\mu$ or $\bar\sigma^\mu$ as appropriate):
$$\mathcal L_\text{kin} = i \bar \psi_L \gamma^\mu \partial_\mu \psi_L + i \bar \psi_R \gamma^\mu \partial_\mu \psi_R $$
Replacing $\psi_{L,R}$ by the fields with the Goldstone pulled out:
$$ \mathcal L_\text{kin} = i \bar \psi_L' e^{-iq \pi(x)/f} \gamma^\mu \partial_\mu \left( e^{iq \pi(x)/f}\psi_L' \right)+ = i \bar \psi_R' e^{iq \pi(x)/f} \gamma^\mu \partial_\mu \left( e^{-iq \pi(x)/f}\psi_R' \right)$$
In addition to the usual kinetic terms, these give terms where the derivative acts on the Goldstone, $\pi(x)$. These are the interaction terms that are our primary focus. For simplicity, let us combine the left- and right handed chiral spinors $\psi_{L,R}'$ into a Dirac spinor, $\Psi = (F',f')^T$ and use the projection operators $\frac{1}{2}(1\pm \gamma^5)$:
$$\mathcal L_\text{int} = i \left(i\frac{q}{f}\partial_\mu \pi \right) \bar\Psi \gamma^\mu \frac{1}{2}\left(1-\gamma^5\right) \Psi + i \left(-i\frac{q}{f}\partial_\mu \pi \right) \bar\Psi \gamma^\mu \frac{1}{2}\left(1+\gamma^5\right) \Psi $$ These terms combine simply into: $$\mathcal L_\text{int} = \frac{q}{f}\left(\partial_\mu \pi\right)\bar\Psi \gamma^\mu \gamma^5 \Psi \ . $$
We thus arrive at the the Goldstone--fermion interaction term in the shift-symmetric manifest form: clearly $\pi$ is invariant under $\pi(x) \to \pi(x) + c$ and so it is protected from quantum corrections that might generate a mass term $m_\pi^2 \pi^2$.
Using the fermion equation of motion
Now we can use the fermion equation of motion to convert this shift-symmetric form of $\mathcal L_\text{int}$ into one that is manifestly pseudoscalar. Recall that the equation of motion in Dirac notation is:
$$i\gamma^\mu\partial_\mu \Psi = m\Psi$$
Armed with this, we may now integrate $\mathcal L_\text{int}$ by parts to shift the derivative from the $\pi$ to the fermion bilinear. We assume that there's no surface term so that integration by parts in the action amounts to a minus sign and moving the derivative in the Lagrangian:
\begin{align} \mathcal L_\text{int} &= \frac{q}{f} \pi \partial_\mu \left(\bar \Psi \gamma^\mu \gamma^5 \Psi \right) \\ & = \frac{q}{f} \pi\left[ (\partial_\mu\bar\Psi)\gamma^\mu\gamma^5 \Psi + \bar\Psi \gamma^\mu \gamma^5 \partial_\mu \Psi \right] \\ & = \frac{q}{f} \pi\left[ (\partial_\mu\Psi)^\dagger \left(\gamma^0\gamma^\mu \gamma^0\right) \gamma^0\gamma^5 \Psi - \bar\Psi \gamma^5 \gamma^\mu \partial_\mu \Psi \right] \\ & = \frac{q}{f} \pi\left[ (\gamma^\mu\partial_\mu\Psi)^\dagger \gamma^0\gamma^5 \Psi - \bar\Psi \gamma^5 \gamma^\mu \partial_\mu \Psi \right] \\ & = \frac{q}{f} \pi\left[ (-im\Psi)^\dagger \gamma^0\gamma^5 \Psi - \bar\Psi \gamma^5 \left(-im\Psi\right) \right] \\ & = \frac{2iqm}{f} \pi \bar\Psi\gamma^5 \Psi \ . \end{align}
This now gives us our manifestly pseudoscalar interaction between the Goldstone $\pi$ and the fermions.
Reiteration of the puzzle
So the puzzle is that:
The shift-symmetric and manifestly pseudoscalar forms of the interaction seem perfectly equivalent.
However, the pseudoscalar form of the interaction seems perfectly general. One could tune the fermion mass $m$ to be whatever you want by tuning the Yukawa coupling $y$. This, in turn, tunes $g = 2qm/f$ to be any pseudoscalar coupling. Does this mean that any pseudoscalar interaction between massive fermions can be written as a Goldstone interaction?
In the manifestly pseudoscalar version of the theory, are loop contributions to the $\pi$ mass manifestly zero? This does not seem to generically be the case. (See, e.g. this discussion based on a problem in Peskin & Schroeder)
So: in the case where there really is a spontaneously broken symmetry, there should be a shift symmetry protecting the $\pi$, but how can we see the effect of that shift symmetry when we calculate loops in the pseudoscalar theory?
Alternatively, if we took a generic pseudoscalar theory with no shift symmetry (i.e. the pseudoscalar is not a Goldstone), then what prevents me from using the equation of motion to write the interaction in a manifestly shift-symmetric form and waving my hands that there ought to be a shift symmetry?