Can the Carmichael Function be used with Proof By Infinite Descent in the following way? I was reading up on the Carmichael Function and I had a question about its use.
Can it be used to show that if $x > 1$, then $n=1$ for:
$$2^m(2^x - 1) = 3^n(3^y - 1)$$
Here's the argument that I came up with.  Did I make a mistake?
Let:

*

*$m,x>1,n,y$ be integers

(1)  Assume that $n > 1$
(2)  $2^x \equiv 1 \pmod {3^n}$
(3)  From the Carmichael Function, $2^{\varphi(3^n)}$ is the least power of $2$ which is congruent to $1 \pmod {3^n}$ with $\varphi(3^n) = 2\times3^{n-1}$
(4)  It follows from Lagrange's Theorem that $2\times3^{n-1} | x$ so that there exists an integer $b$ such that:
$$2^m(2^x - 1) = 2^m(2^{2\times3^{n-1}b} - 1) =2^m(2^{3^{n-1}b}-1)(2^{3^{n-1}b}+1) = 3^n(3^y - 1)$$
(5) Since $n > 1$, it follows that $2\times3^{n-1} | (3^{n-1}b)$ and there exists $c \ge 1$ such that $2^{3^{n-1}b}-1 = 2^{3^{n-1}2c}-1 = (2^{3^{n-1}c} - 1)(2^{3^{n-1}c} + 1)$.
(6) Indeed, it appears to me that we have an example of infinite descent since we can continue to repeat this same process without end since in each case $n > 1$ and this power must therefore be divisible by $\varphi(3^n)$
(7)  Therefore, we can reject our assumption at step(1).
 A: There are several issues with your argument. First, with

(3)  From the Carmichael Function, $2^{\varphi(3^n)}$ is the least power of $2$ which is congruent to $1 \pmod {3^n}$ with $\varphi(3^n) = 2 \times 3^{n-1}$

The Carmichael function, i.e., $\lambda(n)$, is the smallest positive integer where all integers $a$ coprime to $n$ have $a^{\lambda(n)} \equiv 1 \pmod{n}$. However, for any specific coprime integer $a$, there may be smaller such exponents, with the smallest positive one being called the multiplicative order (note the Carmichael function value is the $\operatorname{lcm}$ of all of the multiplicative orders). Nonetheless, since $2$ to any odd power will be congruent to $2$ modulo $3$, the multiplicative order must be even, so since it must divide $\varphi(3^{n})$, it will be $2 \times 3^{k}$ for some $0 \le k \le n - 1$. Also, as I recall, there was a question fairly recently on this site which was basically equivalent to proving that the multiplicative order, in this case, is actually $2 \times 3^{n - 1}$.
Assuming this is correct, the next issue is with

(5) Since $n > 1$, it follows that $2\times3^{n-1} | (3^{n-1}b)$ and there exists $c \ge 1$ such that $2^{3^{n-1}b}-1 = 2^{3^{n-1}2c}-1 = (2^{3^{n-1}c} - 1)(2^{3^{n-1}c} + 1)$.

From your $(4)$, you got that $2^m(2^{3^{n-1}b}-1)(2^{3^{n-1}b}+1) = 3^n(3^y - 1)$. Your step $(5)$ is correct only if $b$ is even. If $b$ is odd instead (note since each repeat of your $(4)$ and $(5)$ removes a factor of $2$ from the exponent, this will happen once all those even factors have been removed), then $2^{3^{n-1}b} - 1 \equiv 1 \pmod{3}$, and $3$ instead divides $2^{3^{n-1}b} + 1$. In this case, you can't split the result into another difference of squares, so you will not get your infinite descent you describe in your $(6)$.
One final issue is I don't see anywhere in your argument that $n \gt 1$ is actually used and makes a difference, i.e., the steps you wrote also seem to apply to the case where $n = 1$ as well, which does have solutions, e.g., $m = 3$, $x = 2$, $n = 1$ and $y = 2$.
Note that I believe your hypothesis is correct, but I'm not sure offhand how to prove it, or even if there is any current method available to do this. Although I only manually checked a few relatively small values, I also suspect that if the hypothesis is not true, the first counter-example may be quite large and difficult to find.
