Let's find all positive integers $n$ for which $2^{n}-1$ and $2^{n+2}-1$ are both prime, and $2^{n+1}-1$ is not divisible by 7.
Solution. The remainders of the powers of 2 when divided by 7 are 1, 2, 4, 1, 2, 4, 1, and so on, repeating periodically. In the problem, three consecutive powers of 2 are involved, and we subtract 1 from each, so their remainders when divided by 7 are definitely 0, 1, 3, in some order. Therefore, there is a number among them that is divisible by 7, but it cannot be $2^{n+1}-1$, so it must be one of the other two. We know that these are prime numbers. If a prime number is divisible by 7, then it can only be 7.

Thus, there are two possible cases: $2^{n}-1=7$ or $2^{n+2}-1=7$.

In the first case, $n=3$, for which 7 is a prime number, $2^{5}-1=31$ is a prime number, and $2^{4}-1=15$ is not divisible by 7, so $n=3$ is a valid solution.

In the second case, $n=1$. Then $2^{1}-1=1$, which is not a prime number, so this is not a solution. Therefore, the only solution is $n=3$.