$T$ be linear operator on $V$ by $T(B)=AB-BA$. Prove that if A is a nilpotent matrix, then $T$ is a nilpotent operator.

This is in the same spirit to your proof, but presented in a different way. If $\lambda B = AB-BA$ for some $B\ne0$ and some $\lambda$ in the algebraic closure of $F$, then $(A-\lambda I)B=BA$ and $(A-\lambda I)^k B=BA^k$ for any $k\ge1$. In particular, $(A-\lambda I)^nB=0$. However, if $\lambda$ is nonzero, $A-\lambda I$ would be invertible and hence $B=0$, which is a contradiction.

This is not an answer as it depends on certain characteristics of the underlying field (also, I haven't used the fact that $A$ is nilpotent).

Suppose $T(B) = \mu B$, that is $\mu B = AB-BA$. Then $\mu B^2 = BAB-B^2A= (AB-\mu B)B-B^2 A= AB^2 -B^2A - \mu B^2$, or $2 \mu B^2 = AB^2 -B^2 A = T(B^2)$.

So, if $ \mu$ is an eigenvalue corresponding to an eigenvector $B$ of $T$, then $B^2$ is an eigenvector corresponding to the eigenvalue $2 \mu$. Hence $2^k \mu$ are eigenvalues.

Since there are only a finite number of eigenvalues (this is where I am making presumptions about the field), we have $\mu = 0$.

Your proof seems correct to me. This result is used for Engel's theorem in the theory of Lie algebras. If $x\in \mathbb{gl}(V)$ is nilpotent, then also $ad(x)$ is nilpotent, where $ad(x)(y)=[x,y]=xy-yx$ for $x,y\in \mathfrak{gl}(V)$. Indeed, $ad(x)^m$ is a linear combination of terms $x^iyx^{m-i}$.