Coupling constants in QFT equations of motion

My question is about the coupling constants that appear in the quantum analogue to the equations of motion in a classical field theory (the Schwinger-Dyson equations). As a concrete example say we have the Lagrangian $$\mathcal{L} ~=~ \frac{1}{2} \left ( \partial_{\mu} \phi \partial^{\mu} \phi - m^2 \phi^2\right ) - \frac{\lambda}{4!} \phi^4.$$ Then by a short argument (replace $\phi$ by $\phi+\epsilon$ in the path integral) we should get the vacuum expectation value of the equations of motion vanishing $$\left \langle (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 \right \rangle ~=~ 0.$$ We can also include other operators inside the expectation value if we include the proper contact terms. So in the following (so that the equation is not trivial) imagine that the equation of motion is multiplying some operator.

Now my question is for which $m$ and $\lambda$ is this true? The bare coupling constants don't make sense without a corresponding regularization scheme and scale $\Lambda$, which does not appear in this equation in an obvious way. Likewise the renormalized coupling constants are associated with a scale.

It seems to me these equations must be ill defined and there must be some regularization at some point. And I see that the product of fields at the same point in $\langle\phi^3(x)\rangle$ is problematic. I was wondering if someone could clarify how this equation is interpreted and how the renormalization of the coupling constants comes about?

Yes, the field equations are ill-defined and must be renormalized. This has been done for simple theories like $\phi^4$ and QED in old work by Brandt and Zimmermann around 1970.