Let $f:M_n(F)\rightarrow M_n(F), X\mapsto AX-XA$. If $f$ is diagonalizable, I want to show that $A$ is diagonalizable. I'd prefer to avoid Jordan Blocks. I know that $f$ is diagonalizable if and only if:
If $F$ is a splitting field for $A$ (e.g. when $F$ is algebraically closed), we may prove the statement as follows.
Proof 1. (A simplified version of user8675309’s answer.) Let $\lambda\in F$ be an eigenvalue of $A$ and $v\in F^n$ be a corresponding left eigenvector. Let $B=A-\lambda I$. Define $g:M_n(F)\to M_n(F)$ by $g(X)=BX$. Since $A$ and $B$ commute, so do $f_A$ and $g$. Moreover, for any vector $x\in F^n$, we have $$ f_A(xv^T)=Axv^T-xv^TA=Axv^T-x(\lambda v^T)=Bxv^T=g(xv^T). $$ It follows that $(m(f_A))xv^T=(m(g))xv^T=m(B)xv^T$ for every polynomial $m\in F[x]$. In particular, when $m$ is the minimal polynomial of $f$, we have $m(B)xv^T=0$. Since $x$ is arbitrary and $v$ is nonzero, we must have $m(B)=0$. Hence the minimal polynomial of $B$ divides $m$. However, as $f_A$ is diagonalisable over $F$, $m$ is a product of distinct linear factors. Therefore the minimal polynomial of $B$ is also a product of distinct linear factors. This means $B$ is diagonalisable over $F$. In turn, $A=B+\lambda I$ is diagonalisable over $F$ too.
Proof 2. Since $F$ is a splitting field for $A$, $A$ admits a Jordan-Chevalley decomposition $S+N$. As $S$ is diagonalisable, so is $f_S$. In fact, if $\{v_1,\ldots,v_n\}$ is an eigenbasis of $S$, then $\{v_iv_j^T: i,j\in[n]\}$ will be an eigenbasis of $f_S$. One may also verify that $f_N$ is nilpotent (with $f_N^{2n-1}=0$), $f_S$ commutes with $f_N$, and $f_A=f_S+f_N$. Therefore $f_S+f_N$ is a Jordan-Chevalley decomposition of $f_A$. By assumption, $f_A$ is diagonalisable. Hence $f_N=0$. In turn, $N=0$. Therefore $A=S$ is diagonalisable.
I assume $\mathbb F$ is algebraically closed (or at least a splitting field for $A$) and $\text{char }\mathbb F \neq 2$.
Let $\lambda$ be an arbitrary eigenvalue for $A$ and $B:= \big(A-\lambda I\big)$. We need to show $\lambda$ is semi-simple $\iff \dim \ker B \leq \dim \ker B^2$ is met with equality. We can re-write $f$ as
$f\big(X\big) = AX - XA = AX - XA - \lambda I X - X (-\lambda I) = BX - XB$
Now suppose for contradiction that $\dim \ker B \lt \dim \ker B^2$
i.e. there is some $\mathbf x \in \ker B^2$ but not in the nullspace of $B$ and some corresponding $\mathbf y^T$ is the left nullspace of $B^2$ but not in the left nullspace of $B$ and set $X:=\mathbf {xy}^T$
(i.) if $X \in \ker f$
$\mathbf 0 = f\Big(f\big(X\Big)\Big)= f^2\big(X\big) = B^2X + XB^2 - 2BXB=-2BXB = -2\big(B\mathbf x\big)\big(\mathbf y^TB\big)$
which contradicts $\mathbf x,\mathbf y^T$ not being in the right and left nullspaces of $B$ respectively
(ii.) if $X \not\in \ker f \implies X \not\in \ker f^3$
since $f$ is diagonalizable
$\mathbf 0 \neq f\Big(f^2\big(X\Big)\Big)= f\big(-2BXB\big)=B\big(-2BXB\big)-\big(-2BXB\big)B =-2B^2XB + 2BXB^2 = \mathbf 0 +\mathbf 0$
which is another contradiction
conclude $\lambda$ is semi-simple and $A$ is diagonalizable
remark on characteristic 2
Part (ii) is even easier and reads
$\mathbf 0 \neq f^3\big(X\big) = f\Big(f^2\big(X\big)\Big)= f\Big(2BXB\Big) = f\Big(0BXB\Big)=\mathbf 0$
Part (i) reads
$\mathbf 0 =f\big(X\big)=BX -XB=BX +XB$
i.e. $BX=XB \neq \mathbf 0$. But $\big(B\mathbf x\big)\big(\mathbf y^T B\big) =BXB =B^2X =\mathbf 0$, generating the same contradiction as before, albeit with the scalar of $2$ removed.
In a comment elsewhere, @Justauser pointed out that, for separably closed fields, your claim follows from the general machinery that algebraic representations carry semisimple (respectively, nilpotent) elements to semisimple (respectively, nilpotent elements). (See Theorem 4.4 of Borel - Linear algebraic groups for the result in the generality in which I have stated it, and Proposition 1.24 of version 2.00 of Milne - Lie algebras, algebraic groups, and Lie groups for the special case of interest here.) $\DeclareMathOperator\ad{ad}\DeclareMathOperator\End{End}$Thus, writing $\ad$ for the algebraic representation $\End_F(F^n) \to \End_F(\End_F(F^n))$ of $\End_F(F^n)$ on itself given by $Y \mapsto (X \mapsto X Y - Y X)$, we have that $\ad(A_\text s)$ is semisimple, $\ad(A_\text n)$ is nilpotent, and (by the Jacobi identity!) they commute, so that $\ad(A) = \ad(A_\text s) + \ad(A_\text n)$ is the Jordan decomposition of $\ad(A)$. In particular, if $\ad(A)$ is semisimple, then $\ad(A_\text n)$ equals $0$, so $A_\text n$ is central in $\End_F(F^n)$; but the centre of $\End_F(F^n)$ consists of the scalar matrices, of which the only nilpotent one is the $0$ matrix. So $A_\text n$ equals $0$, and hence $A = A_\text s$ is semisimple. Over a separably closed field (or, as, for example, @user1551 observed, even just a field that contains all eigenvalues of $A$), this implies that $A$ is diagonalisable.
In general, we seem to be able to conclude only that $A$ is semisimple, not necessarily diagonalisable; but let's dig a bit deeper. We have that the eigenvalues of $\ad(A)$ are the differences of eigenvalues of $A$ (see, for example, @Justauser's answer to What are the eigenvalues and eigenvectors of $\operatorname{ad}_x$ for non-diagonalizable $x$?); so, if $\ad(A)$ is diagonalisable, then every pair of eigenvalues of $A$ differ by an element of $F$. In particular, since $A$ is $F$-rational, so that its eigenvalues are closed under the action of the absolute Galois group of $F$, we have for every eigenvalue $\lambda$ of $A$ and every $\sigma$ in the absolute Galois group of $F$ that $\sigma(\lambda) - \lambda$ belongs to $F$, so that, if $m$ is the order of the restriction of $\sigma$ to the Galois closure of $F[\lambda]$, then $\sum_{i = 0}^{m - 1} \sigma^i\left(\sigma(\lambda) - \lambda\right)$ equals both $\sigma^m(\lambda) - \lambda = 0$ and $m(\sigma(\lambda) - \lambda))$. If $F$ has characteristic $0$, or, more generally, if the Galois closure $E$ of every extension of $F$ of degree at most $n$ satisfies $[E : F] \ne 0$ in $F$, then this implies that $\sigma(\lambda) - \lambda$ is $0$. Since $\sigma$ was an arbitrary element of the Galois group, this implies that $\lambda$ belongs to $F$. Since $\lambda$ was an arbitrary eigenvalue of $A$ (and we already know that $A$ is semisimple), this implies that $A$ is diagonalisable.
If $F$ has characteristic $p > 0$, then $A$ can fail to be diagonalisable. For one family of examples, suppose that $F$ admits a cyclic, Galois extension $E$ of degree divisible by $p$. Fix a generator $\sigma$ of $\operatorname{Gal}(E/F)$. By the additive version of Hilbert's theorem 90 (see, for example, Theorem 6.3 of Lang - Algebra), there is an element $\lambda \in E$ such that $\sigma(\lambda) - \lambda$ is a non-$0$ element of $F$. Then the operator $A$ of multiplication by $\lambda$ on $E$ is an element of $\End_F(E)$ such that $\ad(A)$ is diagonalisable, but $A$ itself is not diagonalisable. For a concrete example of this, you might consider the cubic extension $E$ of $F = \mathbb F_3$ obtained by adjoining a root $\lambda$ of the polynomial $X^3 - X - 1 \in F[X]$, irreducible because it is cubic and has no roots in $F$, with $\sigma$ being the cubing map on $E$. Then, by construction, $\lambda^3 - \lambda$ equals $1 \in F$.
