Given some $n\times n$ matrices $ A_{i},i=1,\dots,m $, over a finite field $\mathbb{F}_p$. Suppose that all $ A_{i} $ are nilpotent and
$$ c_{i}A_{i}+\sum_{j\neq i}b_{ij}[A_{i},A_{j}]=0 $$
for some $ c_{i}\neq 0,b_{ij}\in \mathbb{F}_{p} $ where $ [A_i,A_j] := A_i A_j - A_j A_i $ for all $i, j$. If $m \geq 3$, assume further that $ b_{ij}=0 $ if $ i,j $ are odd. Is it true that $ A_{i}=0 $ for all $ i $? If not, is it possible to give some conditions for the coefficient $b_{ij}$ so that $ A_{i}=0 $ for all $i$?
If $m=2$, then we may assume that $A_1=[A_1,A_2]$ without loss of generality. Then $A_1,A_2$ are simultaneously triangularizable and we can assume that they are strictly triangular matrices. From this, it's easy to see $A_1=0$ and we are done. The case of $m=3$ can be reduced to the cae of $m=2$ since $b_{13}=0$. In general, I tried to prove that they are also simultaneously triangularizable, but I don't know how to prove it.
Note that if we drop the condition "$ b_{ij}=0 $ if $ i,j $ are odd", then the claim is not true for $m=3$: take $$A_1=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad A_2=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \qquad A_3=\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$$ and $c_1=c_2=c_3=2$, $b_{12}=-1,b_{13}=1,b_{21}=b_{23}=b_{31}=b_{32}=-1$.