Croatia Mathematical Competitions

Assume that such $n$ exists.
Note that $5$ divides $6^n - 1 = (6-1)(6^{n-1} + \dots + 1)$, so $5$ also has to divide $7^n - 1$.
The powers of $7$ when divided by $5$ give remainders $2, 4, 3, 1, \dots$ and these remainders repeat periodically. Therefore, $7^n - 1$ will be divisible by $5$ if and only if $n$ is divisible by $4$, i.e. $n = 4k$.
By the formula for the difference of $k$th powers, we can conclude that $6^4 - 1$ divides $6^{4k} - 1$. Moreover, $7$ divides $6^4 - 1 = (6^2 - 1)(6^2 + 1) = 35 \cdot 37 = 5 \cdot 7 \cdot 37$. Therefore, $7$ divides $6^n - 1$, so it also divides $7^n - 1$, which is a contradiction.