Do we have $$ 7^k\mid 3^n+5^n-1\Longleftrightarrow n=t\cdot 7^{k-1}\quad\text{with}\quad t\equiv 1,5\,(\operatorname{mod}6), $$ where $k$ is an arbitrary positive integer?
The part "$\Leftarrow$" is easy. Let $a=3^{t\cdot 7^{k-1}}$, then $5^{t\cdot 7^{k-1}}\equiv a^{-1}\,(\operatorname{mod}7^k)$, and $$7^k\mid a^6-1=(a-1)(a+1)(a^2+a+1)(a^2-a+1).$$ We have $a\equiv\begin{cases}3,\text{if }t\equiv 1\,(\operatorname{mod}6)\\5,\text{if }t\equiv 5\,(\operatorname{mod}6)\end{cases}$. Since $3$ and $5$ are a primitive roots modulo $7$, none of $a-1$, $a+1$ and $a^2+a+1$ is divisible by $7$, so $7^k\mid a^2-a+1$, or $3^{t\cdot 7^{k-1}}+5^{t\cdot 7^{k-1}}-1\equiv a+a^{-1}-1\equiv 0\,(\operatorname{mod}7^k)$.
What about the $\Rightarrow$ part? I think it is true, but how to show that $n$ must be divisible by $7^{k-1}$ (which solves the problem by simply looking at $n\operatorname{mod}6$)? Thank you for any help in advance.
My approach (inspired by Ibrahim_K's post here):
Lemma. Let $n>0$ and $0\le m\le n$, then $\nu_p(\binom{n}{m})\ge\nu_p(n)-\min\{\nu_p(m),\nu_p(n-m)\}$, where $\nu_p$ is the $p$-adic valuation.
Proof. Let $d:=\min\{\nu_p(m),\nu_p(n-m)\}$, then $\nu_p(n)\ge d$. There is nothing to prove if $\nu_p(n)=d$. If $\nu_p(n)>d$, then we must have $\nu_p(m)=\nu_p(n-m)=d$, so the $d$-th digits of the base-$p$ expansions of $m$ and $n-m$ are both nonzero, while the $d$-th to $(\nu_p(n)-1)$-th digits of $n$ are all $0$s, so there are carries at least on the $d$-th to $(\nu_p(n)-1)$-th digits when adding $m$ and $n-m$. The result follows from Kummer's theorem.
Now we study $\nu_7(3^n+5^n-1)$. By looking at $(3^n+5^n-1)\operatorname{mod}7$ we see that $7\mid 3^n+5^n-1$ if and only if $n\equiv 1,5\,(\operatorname{mod}6)$, so we will assume that henceforth; in particular $7\nmid 2^n-1$. Then we have $$ \nu_7(3^n+5^n-1)=\nu_7((3^n+5^n+(-1)^n)(2^n-1))=\nu_7((6^n-(-1)^n)+(10^n-3^n)-(5^n-(-2)^n)). $$ Note that $6-(-1)=10-3=5-(-2)=7$. We have $(a+7)^n - a^n = \sum^{n}_{i=1}\binom{n}{i}7^ia^{n-i}$. By the lemma above we know that $\nu_7(\binom{n}{i}7^i)\ge\nu_7(n)+i-\nu_7(i)$. If $i\ge 2$, then $i-\nu_7(i)\ge 2$, so in fact $$ (a+7)^n - a^n = (\text{multiple of }7^{\nu_7(n)+2})+7na^{n-1}. $$ As a result, $$ (6^n-(-1)^n)+(10^n-3^n)-(5^n-(-2)^n)=(\text{multiple of }7^{\nu_7(n)+2})+7n(3^{n-1}+(-1)^{n-1}-(-2)^{n-1}). $$ To conclude that $\nu_7(3^n+5^n-1)=\nu_7(n)+1$, it suffices to show that $$ 7\nmid3^{n-1}+(-1)^{n-1}-(-2)^{n-1}, $$ which is quite direct. We see that the equivalence conjectured in the original question is true.