This is overkilled by Mason-Stothers theorem?

The official solution does start with Official solution wrote: This conclusion is a consequence of the following version of the Mason-Stothers theorem: , but are there any other solutions?

Any elementary solution only by observing the complex root without Mason Stothers?

Let $A^m - B^n = P$ be a polynomial with degree less than $\min(m,n)$ by condition. Therefore, we know that \[ m \ \text{deg} A = n \ \text{deg} B \]Now, taking derivatives, we have \[ m \cdot A^{m - 1} \cdot A' - n \cdot B^{n - 1} \cdot B' = P'\]\[ m \cdot A^m \cdot A' - n \cdot B^{n - 1} \cdot AB' = AP' \]\[ m(B^n + P)A' - n \cdot B^{n - 1} \cdot AB' = AP' \]Thus, \[ B^{n - 1}(mBA' - nAB') = AP' - mPA' \]Therefore, $B^{n - 1} | AP' - mPA' $, observing degree we have \[ (n - 1) \text{deg} B \le \min\{m,n\} + \text{deg} A - 1 \]Which is then impossible. Therefore, we must have $P = 0$.

Wow. Same exact trick as USA TST 2017/3 and RMM 2018/2. Suppose $A^m-B^n=C$ where $C\ne 0$ but $\deg C<\min\{m,n\}$. We take a derivative to get \[mA^{m-1}A'-nB^{n-1}B'=C'.\]By combining the equations, we get \[mCA'-C'A = B^{n-1}[nB'A-mBA']\]and \[nB'C-BC'=A^{m-1}[nB'A-mBA'].\]We now do some degree counting. ( ) Since $A^m-B^n$ has lower degree than the top terms of both $A^m$ and $B^n$, we see that their top terms must have canceled, so $m\deg(A)=n\deg(B)$. Note that the top term of $mCA'-C'A$ has degree $\deg(A)+\deg(C)-1$ and coefficient \[m\cdot\deg(A)-\deg(C)>0,\]which is thankfully nonzero. Thus, $\deg(mCA'-C'A)=\deg(A)+\deg(C)-1$. Note that if $nB'A-mBA'$ was $0$, then $m(A'/A)=n(B'/B)$, so $A^m=k\cdot B^n$. Plugging into $A^m-B^n=C$ then gives $C=(k-1)B^n$, so to have $\deg C<\min\{m,n\}$, we must have $C=0$, which is a contradiction. Thus, $nB'A-mBA'\ne 0$. Looking now at the equation \[mCA'-C'A = B^{n-1}[nB'A-mBA'],\]we have that \[(n-1)\deg B\le \deg(A)+\deg(C)-1.\]We also derive in a similar fashion that \[(m-1)\deg A\le \deg(B)+\deg(C)-1.\]WLOG, suppose $m\ge n$. Then, from the first inequality, we get \[\deg(A)\cdot\left[(n-1)\frac{m}{n}-1\right]\le \deg(C)-1\le n-2,\]so \[\deg(A)\le\frac{n-2}{(n-1)\frac{m}{n}-1}\le\frac{n-2}{n-2}=1.\]Thus, all the inequalities must be tight, so $(n-1)\deg(B)=\deg(A)+\deg(C)-1$, $m=n$ and $\deg(A)=1$. Thus, $(n-1)\deg(B)=\deg(C)$, so we must have $\deg(B)=1$ in order to have $\deg(C)\le n-1$. Thus, $A$ and $B$ are both linear, which is the desired contradiction.

$C = A^m - B^n$. We suppose $C \neq 0$. $m \cdot deg(A) = n \cdot deg(B)$. We take 2 case: 1) $m = n$. Then $deg(A) = deg(B)$ and $deg(A) \ge 2$. We apply Mason-Stothers for C, $B^n$, $- A^m$. So we obtain $max \{ deg(A^m), deg(B^n), deg(C) \} \le n_0(A^m B^n C) - 1 \le deg(A) + deg(B) + deg(C) - 1 \le 2 deg(A) + m - 1 - 1 = 2 deg(A) + m -2$. So $(m- 2) \cdot deg(A) \le m - 2$ so $deg(A) \le 1$ which is a contradiction. 2) $m < n$. Then $deg(A) > deg(B)$ and $deg(A) \ge 2$. We apply Mason-Stothers for C, $B^n$, $- A^m$. So we obtain $max \{ deg(A^m), deg(B^n), deg(C) \} \le n_0(A^m B^n C) - 1 \le deg(A) + deg(B) + deg(C) - 1 < 2 deg(A) + m - 1 - 1 = 2 deg(A) + m -2$ so $deg(A) < 1$ which is a contradiction. So $ A^m = B^n$.