Obviously, $6^n -1\equiv 0 \pmod{5}\forall n\in \mathbb N$ (by LTE, factorization, etc). Thus, $5\mid 7^n -1$, which is equivalent to $ 7^n\equiv 2^n\equiv 1 \pmod{5}$. We can easily get that $ord_5(2)=4$, which means that $4\mid n \Longleftrightarrow n=4k$. Thus, $1296^k-1\mid 7^{4k}-1$. By LTE or factorization, we get that $7 \mid1296^k-1\forall k\in \mathbb N$, hence $7\mid 7^{4k}-1$, which means that there are no solutions.