Solving Diophantine Equations Involving Exponents Find all positive integers $a, b, c > 1$, such that $(a, b)=(b, c)=(c, a)=1$ $b\mid 2^{a} +1, c\mid 2^{b} +1, a\mid 2^{c} +1$. $$$$ It is easy to see that all of $a, b, c$ are odd. Suppose $3\mid a$ then we have $3\mid 2^c+1$ and as $(a, c)=1$ so we have $c=6k+1$ or $c=6k+5$ for some $k$. But $9\mid a$ implies $9\mid 2^c+1$ which is not possible. This is true for all the three $a, b, c$. Hence either all of them are equal to $3$ or have some prime factor other than $3$. Suppose the later one is true.  Let $p$ be a smallest prime factor of $b$ other than $3$ then we have $$2^{2a} \equiv 1 \mod p$$ and also we have $$2^{p-1} \equiv 1 \mod p$$
So we have $$2^d \equiv 1 \mod p$$ where $d=\gcd (2a, p-1)$. Now $d \neq 2$ as $p \neq 3$. Also $d \neq 3$ because that would have implied $p=7$ and $3|a$ and looking modulo $7$ we arrive at a contradiction. So let $q$ be a prime factor other than $3$ dividing both $a$ and $p-1$ then we have $q<p$ and also as $a$ divides $2^c+1$ we have $$2^{2c} \equiv 1 \mod  q$$ and also $$2^{q-1} \equiv 1 \mod q$$. Again by the same reasoning as above we can show that there is a prime factor $r$ other than $3$ dividing both $c$ and $q-1$ and hence we have $r<q$ and as $c|2^b+1$ we have $$2^{2b} \equiv 1 \mod r$$ and also we have $$2^{r-1} \equiv 1 \mod r$$. Again by the same reasoning we can show that there exists a prime $s$ other than $3$ dividing both $b$ and $r-1$ and hence $s<r<q<p$ and $s$ divides $b$ bot as $p$ was the smallest prime factor other than $3$. So a contradiction and hence there do not exist such positive integers.
$$$$Is The Above Reasoning Correct
 A: Your reasoning is basically correct, but there are a few fairly minor issues. First, you don't seem to directly use that the values are relatively prime, i.e., $(a, b)=(b, c)=(c, a)=1$. For example, you assume $3 \mid a$ and then later wrote

This is true for all the three $a, b, c$. Hence either all of them are equal to $3$ or have some prime factor other than $3$. Suppose the later one is true.

All of them can't be equal to $3$. Also, you can just state if $a = 3$, then $b \mid a^3 + 1 = 9$, which is not possible, to show $a$ must have other prime factors than $3$, as well as both $b$ and $c$. A second, among several other, example of this issue is where you wrote

Let $p$ be a smallest prime factor of $b$ other than $3$ ...

Since $3 \not\mid b$, you don't need to state "other than $3$".
Another minor issue is in

Also $d \neq 3$ because that would have implied $p=7$ and $3|a$ and looking modulo $7$ we arrive at a contradiction.

You perhaps should have added something like $2^3 \equiv 1 \pmod{7}$ so $2^a + 1 \equiv 2 \pmod{7}$ as an explanation for the contradiction to show why $7 \not\mid b$.
One final issue is your proof technically just proves that $3$ can't be a factor of any of $a$, $b$ or $c$. You can finish the proof properly by stating that repeating the reasoning used above with $3$ not being a factor in any variable will still reach the same contradiction, thus showing there are no solutions.
