A matrix cannot be expressed as a sum $B+C$ with $B^2=B$ and $C^3=0$

Not sure if this is useful or not, but $operatorname{rank}(B)$ must be equal to $4$.

Let $K=Aoplus A$. Since $K$ has zero trace, $B$ has an even rank. Clearly $Bne0$ or $I_8$. Therefore the rank of $B$ can only be $2,4$ or $6$.

Suppose $operatorname{rank}(B)=2$. Since $C^3=0$, the largest Jordan block in the Jordan form of $C$ is at most $3times3$. As $C$ is $8times8$, its Jordan form has at least three Jordan blocks. Therefore $dimker(C)ge3$, i.e. $operatorname{rank}(C)le5$. But then $operatorname{rank}(K)=operatorname{rank}(B+C)leoperatorname{rank}(B)+operatorname{rank}(C)le7$, which is a contradiction to the assumption that $K$ is nonsingular.

Therefore $operatorname{rank}(B)ne2$. Since the characteristic polynomial of $A$ is $x^4+x^3+1$, $1$ is not an eigenvalue of $A$. Hence $K+I$ is a traceless nonsingular matrix, $B+I$ is idempotent and $K+I=(B+I)+C$. So, by applying our previous argument to $(K+I,B+I,C)$ instead of $(K,B,C)$, we also have $operatorname{rank}(B+I)ne2$, i.e. $operatorname{rank}(B)ne6$. Therefore the rank of $B$ must be $4$.