Property of $2^n+1=xy$ I was wondering if the following were true. It makes sense but I'm having trouble concocting any formal reasoning.

Let $2^n+1=xy$ for some integers $x,y>1$ and $n>0$. For $a\in\mathbb{Z}^+$, does $2^a\mid (x-1)$ $\iff$ $2^a\mid (y-1)$?

Without loss of generality, one only needs to prove one direction. However, I'm not sure how to approach the problem. Here's my attempt:
 Suppose $2^a\mid (x-1)$. Then $x\equiv 1$ mod $2^a$, so
$$y\equiv xy=2^n+1\equiv 2^r+1 \hspace{5mm}(\text{mod }2^a),\hspace{5mm}\text{ where $0\leq r<a$}.$$
I'm having trouble continuing from here, since I don't know any extra information to determine $2^r\equiv 0$ mod $2^a$. It doesn't seem like that could be true for arbitrary $a\in\mathbb{Z}$, so I must have veered horribly off-track. I appreciate any help!
 A: I'll assume $a \ge 1$ and $2^a \mid (x-1)$.
$2^n$ is zero mod $2^a$.  This is because $a < n$: if not, you would have $2^n \mid 2^a$, so $2^n \mid (x-1)$.  In particular, $2^n \le x-1$, which can't hold, as $2^n = xy - 1 > x - 1$.
Therefore, your argument gives $y = 2^n + 1 = 1 \mod 2^a$.
A: *

*If $a\le n$ and $xy=2^n+1$, then $x\equiv1$ implies $$\begin{array}{c l} y & \equiv yx \\ & \equiv 2^n+1 \\ & \equiv 2^{n-a}(2^a)+1 \\ & \equiv2^{n-a}(0)+1 \\ & \equiv 1 & \mod 2^a  \end{array}$$ and hence $2^a\mid(x-1)\implies 2^a\mid(y-1)$. The reverse implication holds by symmetry (simply interchange the roles of $x$ and $y$), so this is in fact a material equivalence.

*If $a>n$ and $xy=2^n+1$, then $2^a>2^n= xy-1> x-1$ and hence $2^a\mid(x-1)$, which means $x=1$, in which case $y\equiv2^n+1\equiv 1\bmod 2^a\iff 2^n\mid 2^a$, and hence is true.

A: You should have $2^n\equiv 0$ mod $2^a$ instead of $2^n\equiv 2^r$ mod $2^a$
A: Let $x=2^ac+1\ and\ y=2^bd+1$  where c,d are odd natural numbers and a≥b>0.
So, $(2^n+1)=xy=(2^ac+1)(2^bd+1)$
=>$(2^n+1)=xy=(2^{a+b}cd+ 2^ac+ 2^bd +1)$
=>$2^n=(2^{a+b}cd+ 2^ac+ 2^bd)$
=>$2^{n-b}=(2^acd+ 2^{a-b}c+ d)$.
=>d=$2^{n-b}-2^acd+ 2^{a-b}c\ $ will be even unless a=b.
So, if $(2^n+1)$ has a factor of the form $2^ac+1$, the other must be of the form $(2^bd+1)$ where c,d are odd natural numbers.
A: Hint $\ $ Case $\rm\, 1\!:\ \, 2^a\mid 2^n,\, \ so\  \ mod\ 2^a\!:\,\ xy \equiv 1+2^n\equiv 1,\ $ so $\rm\,\ x\equiv 1\iff y\equiv 1.\ $ 
Case $\rm\,2\!:\ \ 2^n\mid 2^a,\,$ so $\rm\:xy\!-\!1 = 2^n\mid 2^a\mid x\!-\!1\:\Rightarrow\:y=1\Rightarrow\Leftarrow\,$ (by $\rm\:xy\!-\!1 > x\!-\!1\:$ for $\rm\:y>1).$
