Does $Ext^1(M,M) \neq 0$ imply $Ext^2(M,M) \neq 0$?

Let $$A$$ be a finite dimensional symmetric quiver algebra. In https://folk.ntnu.no/oyvinso/Papers/symmetric.pdf corollary 2.6. it was noted that for a simple module $$M$$, we have $$Ext^1(M,M) \neq 0$$ implies $$Ext^2(M,M) \neq 0$$. Now I noted that for a general indecomposable module $$M$$ over a representation-finite symmetric algebra $$Ext^1(M,M) \neq 0$$ implies $$Ext^2(M,M) \neq 0$$ and for Brauer tree algebras $$Ext^1(M,M) \neq 0$$ even implies $$Ext^i(M,M) \neq 0$$ for all $$i >1$$ (which is not true for general representation-finite symmetric algebras).

Question: I used the classification of representation-finite symmetric algebras up to stable equivalence to prove this (it was rather ugly). Is there a proof not using the classification, that might even work for general Artin algebras (not just quiver algebras)?

The result is not true for general symmetric algebras (see for example Liu-Schulz example algebras), but in case my question has a positive answer, maybe one expect this to hold for a slightly more general class than representation-finite symmetric algebras.

Bonus question: Let $$A=kG$$ be a group algebra and let $$M$$ be an indecomposable $$A$$-module. Does $$Ext_A^1(M,M) \neq 0$$ imply $$Ext_A^2(M,M) \neq 0$$ or even $$Ext_A^i(M,M) \neq 0$$ for all $$i>0$$? This is true in case $$A$$ is representation-finite.