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 ?

First of all $x$ can't bigger than $y$, as then obviously $3^x > 1 + 2^y$. 
Another way to solve the question is to note that if $y \ge 2$ then $x$ is even, because $4 \mid 2^y = 3^x - 1$. Therefeore if $x=2k$ we have:
$$2^y = 3^{2k} - 1 = (3^k - 1)(3^k + 1)$$
But now $\gcd(3^k - 1, 3^k + 1) = 2$ and from the above equation we must have $3^k - 1 = 2$ and $3^k + 1 = 2^{y-1}$. From the first equation we can conclude $k=1$ and so $x=2$ and plugging in the second one we get $y=3$. These are the only possibilities for $y \ge 2$
Now look at the case $y=1$ separately and you will obtain another solution, namely $x=1; y=1$
