$ \newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)} $
Prove that, if $\gcd(a,b)=1$ and $a>b>0$, then $$ \gcd(a^m - b^m, a^n - b^n) = a^{\gcd(m, n)} - b^{\gcd(m,n)}, \quad 0 \leq m < n . $$ (All variables are integers.)
Let $d>0$ be an integer so that $d = \gcd(a^m - b^m, a^n - b^n)$ and therefore, by definition, $d \backslash (a^m - b^m)$, $d \backslash (a^n - b^n)$.
By definition of modular congruence, if $m > 0$, then $a \equiv b \Mod{m}$ if and only if $m \backslash (b - a)$, we get \begin{equation*} b^m \equiv a^m \Mod{d} \; , \quad b^n \equiv a^n \Mod{d} \; . \end{equation*} According to the Bézout's lemma, there must exist integers $x$ and $y$ so that $mx + ny = \gcd(m, n)$.
We can raise both sides to the power of $x$ or $y$: \begin{equation*} (b^m)^x \equiv (a^m)^x \Mod{d} \; , \quad (b^n)^y \equiv (a^n)^y \Mod{d} \; . \end{equation*} We can "combine" and simplify them as following: \begin{gather*} (b^m)^x \cdot (b^n)^y \equiv (a^m)^x \cdot (a^n)^y \Mod{d} \; , \\ b^{mx + ny} \equiv a^{mx+ny} \Mod{d} \; , \end{gather*} and further as \begin{equation*} b^{\gcd(m, n)} \equiv a^{\gcd(m,n)} \Mod{d} \;. \end{equation*} If follows that \begin{equation} d \backslash (a^{\gcd(m,n)} - b^{\gcd(m, n)}) \; . \tag{1} \end{equation} Moreover, we can factor \begin{gather*} a^n - b^n = (a^{\gcd(m,n)} - b^{\gcd(m,n)})(a^{n-\gcd(m,n)}b^0 + a^{n-\gcd(m,n)-1}b^1 + \ldots + a^0 b^{n-\gcd(m,n)}) \;, \\ a^m - b^m = (a^{\gcd(m,n)} - b^{\gcd(m,n)})(a^{m-\gcd(m,n)}b^0 + a^{m-\gcd(m,n)-1}b^1 + \ldots + a^0 b^{m-\gcd(m,n)}) \; . \end{gather*} Here we can see that $(a^{\gcd(m,n)} - b^{\gcd(m,n)}) \backslash (a^n - b^n)$, $(a^{\gcd(m,n)} - b^{\gcd(m,n)}) \backslash (a^m - b^m)$, and therefore \begin{equation} (a^{\gcd(m,n)} - b^{\gcd(m,n)}) \backslash \gcd(a^m - b^m, a^n - b^n) \;. \tag{2} \end{equation} (1) and (2) together imply \begin{equation*} \gcd(a^m - b^m, a^n - b^n) = a^{\gcd(m, n)} - b^{\gcd(m,n)} \; . \end{equation*}
