Find matrices $A,B$: exists $X,Y$: $AX-BY=[A,B]$ and $XA-YB=[A,B]$, here $X,Y$ are also matrices Question Find matrices $A,B$: such that there exists matrices $X,Y$:  $AX-BY=[A,B]$ and $XA-YB=[A,B]$, here $X,Y$ are also matrices, [A,B]=AB-BA - commutator. 
"Find" means "say something worth" any help is welcome. In particular it is not clear for me: 
0) Are there some nice families of solutions except $[A,B]=aA +bB$? 
1) What is dimension of variety $A,B$?
2) Is it reduced? If not how many components this variety has? 
... and further in future one might be asking algebraic-geometric questions about is the corresponding ideal radical ? normality, singularities... 

Remarks. For fixed $A,B$ we have linear system of equations on $X,Y$.
however this system is always degenerate. For generic matrices $A,B$ it would not have solutions.
See also these questions:
$\det(A \otimes B - B \otimes A) = 0$ why? Why $rk(M) = n^2-n$ ? Why  x and -x in Spec(M) ?  
https://mathoverflow.net/questions/97036/relaxing-commutativity-for-c1-c2-find-q1-q2-1-c1-c2q1c2-q2c1-2-q1-q20/
 A: If $A$ and $B$ are invertible, the following condition is necessary. If $A$ and $B$ are invertible and $A B^{-1}$ has distinct eigenvalues, it is necessary and sufficient:
For all integers $n$, we have
$$Tr{\Large (}[A,B] (AB^{-1})^k {\Large )} = Tr{\Large (}[A,B] (B^{-1} A)^k {\Large )}. \quad (\ast)$$
Proof of necessity: I'll write out the case of $k$ positive; $k$ negative is similar. The left hand side of $(\ast)$ is
$$Tr((AX-BY) (A B^{-1})^k) = Tr (X (AB^{-1})^k A - Y (AB^{-1})^k B) = Tr (XAB^{-1} A B^{-1} \cdots A - Y A B^{-1} \cdots A)$$
where we have used the cyclic symmetry of trace. Similarly, the right hand side of $(\ast)$ is
$$Tr((XA-YB) (B^{-1} A)^k) = Tr(X A (B^{-1} A)^k - Y B (B^{-1} A)^k) = Tr (XAB^{-1} A B^{-1} \cdots A - Y A B^{-1} \cdots A).$$
The two expressions are identical. $\square$
Proof of sufficiency: Let's study the more general question of, given matrices $P$ and $Q$, when there exist $X$ and $Y$ with $AX-BY=P$ and $XA-YB=Q$. 
Let $V \subseteq \mathrm{Mat}_n^2$ be the set of pairs $(P,Q)$ which can be written in this way.
Clearly, $V$ is a linear space, so it is defined by some linear relations. Every linear function on $\mathrm{Mat}_n^2$ is of the form $(P,Q) \mapsto Tr(PT-QU)$ for some matrices $T$ and $U$. Let's figure out for which $(T,U)$ the function $Tr(PT-QU)$ vanishes on $V$.
This occurs if and only if 
$$Tr((AX-BY) T -(XA-YB)U) = Tr (X (TA-AU)) - Tr (Y (TB-BU))=0.$$
for all $X$, $Y$. This occurs if and only if $TA=AU$ and $TB=BU$. 
So we deduce the more general, but less explicit, condition:

Your equation is solvable if and only if we have $Tr([A,B] T) = Tr([A,B] U)$ for all $T$ and $U$ such that $TA=AU$ and $TB=BU$.

Now suppose $A$ and $B$ are invertible. Then $T A B^{-1} = A U B^{-1} = A B^{-1} T$. So $T$ commutes with $A B^{-1}$ and $U$ can be computed from $T$ as $U=A^{-1} T A$. With the additional hypothesis that $A B^{-1}$ has distinct eigenvalues, the space of matrices that commute with $A B^{-1}$ is spanned by the powers of $A B^{-1}$. $\square$.
Some comments on dimension: 
Note that the sequence $(A B^{-1})^k$ satisfies a linear recurrence of length $n$, by the Cayley-Hamilton theorem. So condition $(\ast)$ for any $n$ consecutive values implies $(\ast)$ for all $k$. Also, condition $(\ast)$ is trivial for $k=0$. So condition $(\ast)$ cuts out a space of codimension $\leq n-1$ in the open set of matrices for which $A$ and $B$ are invertible and $A B^{-1}$ has distinct eigenvalues. The space of solutions to your equation thus has dimension $\geq 2 n^2 - (n-1)$.
I didn't check carefully, but I think that the condition $[A,B] \in \mathrm{Span}(A,B)$ has dimension $n^2 + n+1$, so much smaller.
A general comment: The set of $(A,B)$ such that this equation is solvable will be constructable, by Chevalley's theorem. But I see no reason to expect that it is closed. Without a result saying that the image is closed, I don't see a way to equip the image with the structure of an algebraic variety. Without this, questions about singularities and more subtle invariants don't make sense.
