The two-endomorphism nilpotency conjecture for Schur-finite objects

Let $A$ be an object with endomorphisms $\pi_1$ and $\pi_2$. Set

$$a:=\operatorname{tr}(\pi_1)=\operatorname{tr}(\pi_1^{\circ i})\quad\text{for all }i,$$

and

$$b:=\operatorname{tr}(\pi_2)=\operatorname{tr}(\pi_2^{\circ j})\quad\text{for all }j.$$

Assume also that

$$\operatorname{tr}(\pi_1^{\circ i}\circ\pi_2^{\circ j})=0$$

for all $i$ and $j$. Let $\nu$ be the hook partition occurring in the definition of the polynomial $y$, and let $\lambda$ be a partition. The two-endomorphism nilpotency conjecture. If $S_\lambda(A)=0$ and $\lambda\not\supset (b+2)^{a+2}$, then

$$y(\lambda\setminus\nu;\alpha_1\pi_1+\alpha_2\pi_2)\not =0$$

as a polynomial in $\alpha_1$ and $\alpha_2$, and consequently $\c{N}(A)$ is nilpotent. This is a proposed strengthening of the theorem proved immediately before it; the source gives numerical evidence but no resolution.