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.