Express $ \operatorname{gcd}\left(5^{m}+7^{m}, 5^{n}+7^{n}\right) $ in terms of $m$ and $n$

For $k$ odd we have that $$5^k+7^k\equiv 0\mod 3$$ and $$5^k+7^k\equiv 5+7\equiv 4\mod 8$$

For $k$ even we have that $$5^k+7^k\equiv 2\mod 3$$ and $$5^k+7^k\equiv 1+1\equiv 2\mod 8$$

From here we have that if $m,n$ are both odd(i.e. if $m+n$ is even), then $\gcd(5^m+7^m,5^n+7^n)=12$. Otherwise, if between $m$ and $n$ there is an odd and an even number(i.e. if $n+m$ is odd), then $\gcd(5^m+7^m,5^n+7^n)=2$.

So, for $\gcd(m,n)=1$, you could express $$\gcd(5^m+7^m,5^n+7^n)=2\cdot 3^{(m+n+1)\%2}\cdot 2^{(m+n+1)\%2}$$ Where $a\%b$ denotes the remainder that we obtain when dividing $a÷b$.

Ps. This is clearly not a favorable answer since it doesn't seem generalizable.

Case $\,a,b = 5,7\,$ below [homogenization of this], which applies $\color{#90f}{\rm E} = $ Euclid's algorithm, e.g. $$(a_1,a_2,b) \overset{\color{#90f}{\rm E}}= (\bar a_1,\bar a_2,b)\,\ \ {\rm if}\,\ \ {a_i\equiv \bar a_i}\!\!\! \pmod{\!b},\ \text{is used in the first line of the proof}$$

Theorem $\ $ If $\, m,n\in\Bbb N,\ $ $(m,n)\!=\!1\!=\!(a,b),\,$ and wlog $\,m \!=\! 1\!+\!2j\,$ is odd, then

$$ d := (a^{\large m}\!+\!b^m,a^{\large n}\!+\!b^{\large n})= (a\!+\!b,\color{#0a0}{(-\!1)^{\large n}\!+\!1}) =\begin{cases} (a\!+\!b,2) \ \ {\rm if}\,\ 2\mid n\\ (a\!+\!b)\quad\ \, {\rm if}\ \ 2\nmid \!n\end{cases}\qquad $$

Proof $\ \bmod d\!:\ b^{-1}\,$ exists by $\,(d,b) \overset{\color{#90f}{\rm E}}= (a^m,a^n,b)=1\,$ by $\,(a,b)=1.\,$ Let $\,c \equiv a/b:= ab^{-1}$. Then $\, {c^{\large m}}^{\phantom{|^|}}\!\!\!\equiv -1\equiv c^{\large n}\Rightarrow c^{\large 2m}\equiv 1\equiv c^{\large 2n}$ so $\,{\rm ord}\, c^{\large 2}$ divides coprimes $m,n$ so is $1,\,$ so $\,\color{#c00}{c^{\large 2}\equiv 1}.\,$ $\,{-}1\equiv c^{\large m}\!\equiv c^{\large\phantom{,}}\!(\color{#c00}{c^{\large 2}})^{\large j}\!\equiv c\,$ $\Rightarrow\,c\!+\!1\equiv 0\,\overset{\times\ b}\Rightarrow\,a\!+\!b\equiv 0\,$ so $\,d \overset{\color{#90f}{\rm E}}= (a\!+\!b,d) \overset{\color{#90f}{\rm E}}= (a\!+\!b,\,\color{#0a0}{d\bmod a\!+\!b})\,$ is as claimed, by $\!\bmod{\,\color{#0a0}{\!a\!+\!b}}\!:\ \underbrace{a^{\large k}\!+\!b^k \equiv b^k(\color{#0a0}{(-1)^{\large k}\!+\!1})}_{\large\ \ \ \color{#0a0}{ a\ \,\equiv\ -b}\ \ \ \ \ }^{\phantom .}\,$ and $\,(d,b^k\color{#0a0}e)=(d,\color{#0a0}e)\,$ by $\,(d,b)=1$.

Remark $ $ We can easily extend the above to the case when $\,m,n\,$ are not coprime.

Corollary $\ $ If $\,(A,B)=1\,$ and $\,M,N\in \Bbb N,\,$ and wlog $\,M/(M,N)\,$ is odd, then

$\quad(A^M\!+\!B^M,A^N\!+\!B^N)\, =\, (A^{(M,N)}\!+\!B^{(M,N)},C),\,\ \begin{cases} C = 2\ \ {\rm if}\ \ 2\mid N/(M,N)\\ C = 0\ \ {\rm otherwise}\end{cases}$

Proof $\ $ Let $\,D = (M,N),\,\ a = A^D,\ b = B^D.\,$ Then $\,(m,n) := (M/D,\:\! N/D) = 1\,$ and

$\quad\begin{align} (A^{M}\!+\!B^{M},A^{N}\!+\!B^{N})\, &=\, {(A^{D\large m}\!+\!B^{D\large m},(A^{D\large n}\!+\!B^{D\large n})}\\[.2em] &=\,{ \ \ \ (a^{\large m}\ + \ b^{\large m},\, \ \ \ \ a^{\large n}\ + \ b^{\large n})}\end{align}\ $ so the Theorem applies.