For which positive integers $a, b, c$ is it true that $2^{a}-1$ is divisible by $b$, $2^{b}-1$ is divisible by $c$, and $2^{c}-1$ is divisible by $a$?
$a=1, b=1, c=1$ satisfies the conditions. We will show that no other solution exists.

If any of $a, b, c$ is 1 (due to symmetry, we can assume it is $a$), then the others must also be 1, since $b \mid 2^{a}-1=1$ implies $b=1$, and finally $c \mid 2^{b}-1=1$ implies $c=1$.

In the following, we assume that $a, b, c \neq 1$. We will prove by complete induction that $a, b$, and $c$ are not divisible by any prime $p$.

Since $2^{a}-1, 2^{b}-1$, and $2^{c}-1$ are odd numbers, $a, b$, and $c$ are not divisible by 2. Suppose that for a prime $p \geq 3$, the statement holds for all primes less than $p$. We will show that $p$ is not a divisor of any of the numbers.

We prove this statement indirectly. Suppose that $p$ is a divisor of one of the numbers, say $a$. Then from $a \mid 2^{c}-1$, it follows that $p \mid 2^{c}-1$. By Fermat's Little Theorem, $p \mid 2^{p-1}-1$. Let $d$ be the smallest positive integer exponent for which $p \mid 2^{d}-1$ holds. Since the remainders of the powers of 2 when divided by $p$ are periodic, this implies that both $c$ and $p-1$ are multiples of $d$. Clearly, $d$ is not 1 (since $2^{1}=2$ is not divisible by any prime to give a remainder of 1), so it has a prime divisor, and this can only be less than $p$ (since it is at most $p-1$). Thus, $c$ is divisible by a prime less than $p$. This contradicts the induction hypothesis, so the original assumption must be false: none of the three numbers can be divisible by $p$. This completes the inductive step.

Therefore, indeed, $a=1, b=1, c=1$ is the only solution.