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We define $T(X) = AX - XB$ for fixed $A,B$. We allow $X$ to be any matrix in $M_n(F)$.
Write down all the eigenvalues of $T$ in terms of the eigenvalues of $A$ and $B$.
I think I saw another question here which said that for $T(X) = AX - XA$, if $u$ is an eigenvector for $A$ and $v$ an eigenvector for $A^T$, then $uv^T$ is an eigenvector for $T$, but I don't know how to prove this nor do I know if it generalizes if we replace one instance of $A$ with $B$.
Suppose that $\lambda_A$ and $u_A$ is an eigenvalue/eigenvector pair of $A$ and $\lambda_B$ and $u_B$ is an eigenvalue/eigenvector pair of $B^T$, then
\begin{align*} T(u_A u_B^T) &= A u_A u_B^T - u_A u_B^T B\\ &=(A u_A) u_B^T - u_A (B^T u_B)^T\\ &=\lambda_A u_A u_B^T - \lambda_B u_A u_B^T\\ &=(\lambda_A - \lambda_B) u_A u_B^T \end{align*}
which means that $\lambda_A-\lambda_B$ and $u_A u_B^T$ form an eigenvalue/eigenvector pair of $T$.
Now if the matrices $A$ and $B^T$ have exactly $n$ eigenvalues/eigenvectors pairs which means that we can have $n^2$ eigenvalues/eigenvectors pairs for $T$ this way (see below), but $T$ cannot have more of those since $T:M_n(F)\rightarrow M_n(F)$ is a transformation from an $n^2$ dimensional space to an $n^2$ dimensional space.
Let $u_{A_1}$ and $u_{A_2}$ be two linearly independents normalized eigenvectors of $A$ and let $u_{B_1}$ and $u_{B_2}$ be two eigenvectors of $B^T$ (they could be the same), then suppose that $u_{A_1} u_{B_1}^T=u_{A_2} u_{B_2}^T$, so we have \begin{align*} u_{A_1} = (u_{A_1} u_{B_1}^T) u_{B_1} = (u_{A_2} u_{B_2}^T) u_{B_1}=(u_{B_2}^T u_{B_1}) u_{A_2} \end{align*} and so $u_{A_1}$ and $u_{A_2}$ are collinear, since they are normalized they are equal which proves that $u_{A_1} u_{B_1}^T\neq u_{A_2} u_{B_2}^T$.
The same argument can be given for $u_{B_1}\neq u_{B_2}$ and $u_{A_1}$ and $u_{A_2}$ arbitrary.
This means that all the eigenvectors of this form are distinct.
The converse of the claim is not true, i.e., not every eigenvector of $T$ is of the from $v_Av_B^T$. To see this, take $A=B=\pmatrix{0&1\\0&0}$. Then $$ T\pmatrix{x_{11} & x_{12}\\x_{21} & x_{22} } = \pmatrix{x_{21}& x_{11}-x_{22}\\0&x_{21}}. $$ It follows that $\lambda=0$ is an eigenvalue with eigenspace spanned by $$ \pmatrix{0&1\\0&0}, \quad \pmatrix{1&0\\0&1}, $$ where the latter eigenvector is not of the particular form (and the eigenspace does not contain $\pmatrix{1&0\\0&0}$)
If you want only the eigenvalues over $\bar{F}$, the algebraic closure of $F$, then it's not difficult. Here, we stack the matrices row by row. cf.
https://en.wikipedia.org/wiki/Kronecker_product
There are invertible $P,Q$ s.t. $P^{-1}AP=S=[s_{i,j}],Q^{-1}B^TQ=T=[t_{i,j}]$ are triangular. Then $(P\otimes Q)^{-1}(A\otimes I-I\otimes B^T)(P\otimes Q)=(P^{-1}AP)\otimes I-I\otimes (Q^{-1}B^TQ)$
$=S\otimes I-I\otimes T=[s_{i,j}I]-[\delta_{i,j}T]$. The last two matrices are triangular and the diagonal of the matrix result is
$diag(s_{1,1}I,\cdots,s_{n,n}I)-diag(T,\cdots,T)$.
Finally, the required spectrum is $(\lambda_i-\mu_j)_{i,j}$. Moreover, the result is valid when $A\in M_n,B\in M_m$ ($m\not= n$). For example, for $n=2,m=3$, the diagonal is:
$diag(s_{1,1},s_{1,1},s_{1,1},s_{2,2},s_{2,2},s_{2,2})-diag(t_{1,1},t_{2,2},t_{3,3},t_{1,1},t_{2,2},t_{3,3})$.
