Suppose we have the following matrix:
$M=\begin{pmatrix}2^5 & 7^5 & -3^5\\2^2 & 7^2 & -3^2\\2^4 & 7^4 & -3^4\end{pmatrix}$
The trace of $M$ is the well -known Fermat-Catalan solution:
$2^5+7^2-3^4=0$
By the Albert-Muckenhoupt theorem, we have
$M=AB-BA$
We also know by FLT that the first and third row sums of $M$ are not zero.
How do you find an $A$ and $B$ that will give you this particular $M$?
On
Poking around on my TI nSpire CX calculator, I came up with a more exact solution for A and B.
$A=\begin{pmatrix}0 & (191a_3{_2}-933230)/221 & 81\\1 & 5(5a_3{_2}-3066)/221 & 0\\0 & a_3{_2} & -27\end{pmatrix}$
$B=\begin{pmatrix}0 & 49 & -9\\0 & -4 & 0\\1 & -43 & 0\end{pmatrix}$
I tested $a_3{_2}$ with some integers, rationals, and even $a_3{_2}=\pi$ .
Always works fine.
EDIT $1$:
Thanks to loup blanc above for the $\mathbb{Z}$ solution of $a_3{_2}=127.$
I have taken the solution even further on the calculator to come up with an exact solution for the following generalized commutator:
$M=\begin{pmatrix}z^r-y^q & y^p & -z^p\\x^q & y^q & -z^q\\x^r & y^r & -z^r\end{pmatrix}$
Notice that unless Beal's Conjecture is false, the upper left hand corner can never be $x^p$ when $p,q,r\geq3$.
The solution for $A$ and $B$ using the same $a_3{_2}$ system is:
$A=\begin{pmatrix}0 & f_1(a_3{_2}) & z^r\\1 & f_2(a_3{_2}) & 0\\0 & a_3{_2} & -z^{p-q}\end{pmatrix}$
$B=\begin{pmatrix}0 & y^q & -z^q\\0 & -x^q & 0\\1 & -(z^qx^r+z^p)z^{-q}& 0\end{pmatrix}$
The $f_1(a_3{_2})$ and $f_2(a_3{_2})$ are extremely long expressions.
EDIT $2$:
Continuing my efforts on the TI nSpire CX calculator, I came up with a much simpler solution for $A$ and $B$ in the original Fermat-Catalan case that does not depend on $a_3{_2}$.
$A=\begin{pmatrix}0 & -18464/5 & 81\\1 & 0 & 0\\0 & 3066/5 & -27\end{pmatrix}$
$B=\begin{pmatrix}0 & 49 & -9\\0 & -4 & 0\\1 & -43 & 0\end{pmatrix}$
To do this I used $\vec{v}=(x,y,z,p,q,r)=(2,7,3,5,2,4)$ in the following general solutions for $A$ and $B$.
$A=\begin{pmatrix}0 & f_1(\vec{v}) & z^r\\1 & 0 & 0\\0 & f2(\vec{v}) & -z^{p-q}\end{pmatrix}$
$B=\begin{pmatrix}0 & y^q & -z^q\\0 & -x^q & 0\\1 & -(z^qx^r+z^p)z^{-q}& 0\end{pmatrix}$
where $f_1(\vec{v})=-((z^rx^q-z^p)z^qx^r+z^{2q}y^r+z^{p+r}x^q+z^qy^px^q-z^{2p})z^{-q}/(x^{2q}+z^q)$
and $f_2(\vec{v})=((z^{q+r}+z^px^q)z^qx^r-z^{2q}y^rx^q+z^{p+q+r}+z^{2p}x^q+z^{2q}y^p)z^{-2q}/(x^{2q}+z^q)$
Notice that both $f_1(\vec{v})$ and $f_2(\vec{v})$ share the same common denominator!
This same method works for finding $A$ and $B$ for the commutator constructed for Darmon-Granville solutions such as the following $\vec{v}=(x,y,z,p,q,r)=(2,13,5,5,3,5)$. The upper left hand corner in this case is populated by $928$ which factors into $29\cdot2^5$ as expected.
We consider the equation $(*) \;AB-BA=M$, where the unknowns are the $18$ entries of $A,B$. Since the function $f:(A,B)\in M_3\times M_3\rightarrow AB-BA\in \{Z\in M_3;tr(Z)=0\}$ is surjective, the derivative of $f$ has rank $8$. That is, the general solution of $(*)$ has $18-8=10$ free parameters. It remains to know what are these $10$ paremeters and how to choose them, depending on whether we want a solution in i) a real algebraic extension of $\mathbb{Q}$ or ii) in $\mathbb{Q}$ or iii) in $\mathbb{Z}$.
When one has Maple, the least tiring is to consider the case i), using the Grobner basis theory. The shape of a Grobner basis lets you guess a choice of parameters; then we give simple values to these parameters. For instance, we obtain the following -in Maple format- in a real algebric extension of $\mathbb{Q}$ of degree $3$.
A= Matrix(3, 3, [[8, 1, (71723588976/3027762184027595)*c[3, 3]^2+(2984407072706366/15138810920137975)*c[3, 3]-14832628576459297/2162687274305425], [0, (49493708273997/5817888290354509901)*c[3, 3]^2+(578526588058455697/8311268986220728430)*c[3, 3]+6757362833635887381/1187324140888675490, -(162059734160351703/488906242479941239530535)*c[3, 3]^2-(13367942213184840085621/4889062424799412395305350)*c[3, 3]-124833447963523709485393/698437489257058913615050], [0, 0, 0]])
B= Matrix(3, 3, [[-(216984194154/8827294997165)*c[3, 3]^2-(8918857908851339/44136474985825)*c[3, 3]-431960743222798959/44136474985825, 0, 0], [(143447177952/3027762184027595)*c[3, 3]^2+(5968814145412732/15138810920137975)*c[3, 3]+39540735624855006/2162687274305425, 0, 0], [-2, (1898166816/1765458999433)*c[3, 3]^2+(77630790402871/8827294997165)*c[3, 3]-5307136376915304/8827294997165, c[3, 3]]])
where $c[3,3]$ is a root of the equation
$593177130c[3, 3]^3+4879962476371c[3, 3]^2+66387355284897c[3, 3]-8152887366719494 = 0$.
EDIT 1. If we choose $c[3,3]\approx 34.57\cdots$, then we obtain
$A\approx \begin{pmatrix}8& 1&-0.0144322399335206621\\0& 8.1079935808833385963& -.27366129513717087073\\0& 0& 0\end{pmatrix}$,
$B\approx \begin{pmatrix}-16802.729776638927233& 0& 0\\31.971135520132958675& 0& 0\\-2& -295.88084599083230019& 34.573449979585872872\end{pmatrix}$.
Now if you want some solution of type iii) (To do what ?...), then follow the advice of Will Jagy: roll up your sleeves.
EDIT 2. Below, the OP found a couple solution in $M_3(\mathbb{Q})$ that depends on a parameter $a_{3,2}$ -I don't know how he did it, but he did it!-
Curiously , the system "mod(221)": $191a_{3,2}-933230=0,5(5a_{3,2}-3066)=0$ has the solution $a_{3,2}=127$. Thus, we obtain an explicit solution over $\mathbb{Z}$.
$A=\begin{pmatrix}0&-4113& 81\\1& -55& 0\\0& 127& -27\end{pmatrix},B=\begin{pmatrix}0& 49& -9\\0& -4& 0\\1& -43& 0\end{pmatrix}$.
EDIT 3. Comment on the OP's EDIT 1. We must not look for an explicit formula; it is very easy to find a solution over $\mathbb{Z}$ for $p>q,r,x,y,z$ given in $\mathbb{N}^*$. For example, let $p=13,q=9,r=12,x=5,y=8,z=10$. Resuming your choice, the unknown is the second column of $A$: $[w,v,u]^T$. Equality $M=AB-BA$ gives two affine relations where $w,v$ depend on $u$. It remains to obtain integers solutions of the previous system. Here a solution is
$u=305711831847240,v=2446234370904,w=156231348839563264$.