Let $\Delta : M \in \mathcal{M}_n(\mathbb{C}) \mapsto AM+MB$ with $A,B \in \mathcal{M}_n(\mathbb{C})$.
Q1) Show that if $A$ and $B$ are nilpotent matrix then $\Delta$ is nilpotent
Q2) What is the nilpotency index of $\Delta$ ?
Q1) $\Delta = f_A+g_B$ with $f_A : M \mapsto AM$ and $g_B : M \mapsto MB$. $f_A$ and $g_B$ commute so we can expand $(f_A+g_B)^p$ with the binomial expansion. Taking $p= \mathrm{Ind}(A)+\mathrm{Ind}(B)$ leads to $\Delta^p = 0$.
Q2) I can say that $\mathrm{Ind}(\Delta) \leq \mathrm{Ind}(A)+\mathrm{Ind}(B)$ ? Is there an equality ?
The first step is showing that the nilpotency index does not change by replacing $A$ and $B$ with similar matrices. To see this, let $S$ and $T$ be invertible. Define $\rho: M\mapsto SMT^{-1}$ and $\tilde{\Delta}: M\mapsto SAS^{-1} M + MTBT^{-1}$. Then by $$ S^{-1}(SAS^{-1}M+MTBT^{-1})T = A S^{-1}MT + S^{-1}MT B, $$ we have $\rho \Delta \rho^{-1} = \tilde{\Delta}$. Thus, the $n^2\times n^2$ matrices for $\Delta$ and $\tilde{\Delta}$ must be similar.
Taking the binomial expansion idea a little further, we can see that $$ (f_A+g_B)^{Ind(A)+Ind(B)-1} = 0. $$ This is because $f_A^r g_B^s$ in the expansion must have $r+s=Ind(A)+Ind(B)-1$, and we cannot have both $r$ and $s$ satisfying $r\leq Ind(A)-1$ and $s\leq Ind(B)-1$.
Now, we replace $A$ and $B$ with their respective Jordan forms. Then $A^{Ind(A)-1}$ and $B^{Ind(B)-1}$ are nonzero with the following properties:
$A^{Ind(A)-1}$ has a nonzero $p$-th row with the sole nonzero entry is $1$ at $k$-th position, and
$B^{Ind(B)-1}$ has a nonzero $q$-th column with the sole nonzero entry is $1$ at $m$-th position.
If $M$ is the elementary matrix for column interchange for $k$-th and $m$-th columns, then $$ A^{Ind(A)-1} M B^{Ind(B)-1} $$ has its entry $1$ at $(p,q)$ position, thus it is nonzero.
Hence, $$ (f_A+g_B)^{Ind(A)+Ind(B)-2} $$ will have a nonzero constant multiple of $f_A^{Ind(A)-1}g_B^{Ind(B)-1}$ in the expansion which is nonzero by the above analysis. All other terms are zero.
Therefore, the nilpotency index of $\Delta$ is $Ind(A)+Ind(B)-1$.