Let $\,p\,$ be the least prime divisor of $\,n$
Observe $\,p\mid(6^n+7^n)(7^n-6^n)\,$ and little Fermat to obtain:
$p\mid 7^{2n}-6^{2n},\, 7^{p-1}-6^{p-1}\iff\, p\mid (7^{2n}-6^{2n},\, 7^{p-1}-6^{p-1})$
By below theorem and $\,(2n,p-1)=2\, $ we get $\,p\mid 7^2-6^2=13$
Theorem: $\,(a,b)=1,\,a>b\,\,\Rightarrow\,\, (a^m-b^m,a^n-b^n)=a^{(m,n)}-b^{(m,n)}$
Proof: Use $\,x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+\cdots+xy^{k-2}+x^{k-1})\,$
and $\gcd$ definition $n\mid a,b\iff n\mid (a,b)$ to prove:
$a^{(m,n)}-b^{(m,n)}\mid a^m-b^m,\, a^n-b^n\iff$
$a^{(m,n)}-b^{(m,n)}\mid (a^m-b^m,a^n-b^n)=: d\ \ \ (1)$
$a^m\equiv b^m,\, a^n\equiv b^n$ mod $d$ by definition of $d$.
Bezout's lemma gives $\,mx+ny=(m,n)\,$ for some $x,y\in\Bbb Z$.
$(a,b)=1\iff (a,d)=(b,d)=1$, so $(a^m)^x,(b^n)^y$ mod $d$ exist.
$a^{(m,n)}\equiv (a^{m})^x(a^n)^y\equiv (b^{m})^x(b^n)^y\equiv b^{(m,n)}\pmod{\! d}\ \ \ (2)$
$(1)(2)\,\Rightarrow\, a^{(m,n)}-b^{(m,n)}=d$