<s>[INST] Solving $3^x = 2^y + 1$ with $x,y \in \mathbb{N}^2$ I consider the following equation
$$ 3^x = 1 + 2^y \tag{$\star$} $$
with $(x,y) \in \mathbb{N}^2$ and $y \geq 3$.
I would like to show that :

$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$

I assume that $3^x \equiv 1 \; [2^y]$. Writing 
$$ 3^x = \sum_{k=0}^{x} C_{x}^{k} 2^k = 2^x + x 2^{x-1} + \ldots + 2x + 1 $$
I have that $3^x \equiv 1 \; [2^y]$ is equivalent to :
$$ 2^x + x 2^{x-1} + \ldots + 2x \equiv 0 \; [2^y]. $$
This is also equivalent to :
$$ 2^{x-1} + x 2^{x-2} + \ldots + x \equiv 0 \; [2^{y-1}]. $$
Then, I am unsure about how to proceed. I would like to assume that $x > y$ and then, it would follow that $2^{x-1} \equiv 0 \; [2^{y-1}]$. Therefore, I would end up with :
$$ x \big( 2^{x-2} + \ldots + 1 ) \equiv 0 \; [2^{y-1}]. $$
Since $\operatorname{gcd}\big( 2^{y-1}, 2^{x-2} + \ldots + 1 \big) = 1$, it follows that $2^{y-1} \mid x$. But this is not the expected result. Where did I make a mistake ?
 [/INST] 
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$

To show this, I would use that, for every pair of positive integers $(a,s)$ where $s$ is odd, there exists an odd integer $t$ such that
$$3^{2^as}-1=2^{a+2}t\tag1$$
(the proof are written at the end of the answer)

Proof for $2^{y-2}\mid x\implies 3^x\equiv 1\pmod{2^y}$
If $2^{y-2}\mid x$, then, from $(1)$, $3^{x}-1$ is divisible at least by $2^y$. $\quad\blacksquare$

Proof for $3^x\equiv 1\pmod{2^y}\implies 2^{y-2}\mid x$
It is easy to see that $4\not\mid 3^{\text{odd}}-1$.
The claim follows from this and $(1)$. $\quad\blacksquare$

Finally, let us prove $(1)$ by induction on $a$.
For $a=1$, we have $3^{2s}-1\equiv 1-1\equiv 0\pmod{2^3}$. Also, writing $s=2u+1$, we have $$3^{2s}-1=9^{2u+1}-1=9\cdot 81^u-1\equiv 9-1\equiv 8\pmod{2^4}.$$
Suppose that $(1)$ holds for some $a$.
Then, we have
$$3^{2^{a+1}s}-1=(3^{2^as})^2-1=(2^{a+2}t+1)^2-1=2^{2a+4}t^2+2^{a+3}t=2^{a+3}t'$$
where $t'$ is odd. $\quad\blacksquare$
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