<s>[INST] If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Is it necessarily the case that $m=1$ and that these  two divisors are $2^k-1$ and $2^k+1$?
I've tested this up to $y\leq10^{10}$ but I haven't been able to make much progress with standard number theoretic techniques.
If $k=1$ then there are infinitely many solutions of the form $x=y-1$.

Let $(1)$ be the initial version of the problem and let $(2)$ be the supposedly equivalent version of the problem.
$(2)\implies(1)$: Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. We can write $y=m(2^k-1)+1$ for some $m\geq1$. Then
$$2^ky-1=2^k(m(2^k-1)+1)-1=(2^k-1)(2^km+1)$$
so $y-x$ and $y+x$ are two positive divisors of $(2^k-1)(2^km+1)$ which average to $y=m(2^k-1)+1$. By $(2)$, $y-x=2^k-1$ and $y+x=2^k+1$. Then $x=1$ and $y=2^k$.
$(1)\implies(2)$: Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two positive divisors of $(2^k-1)(2^km+1)$ which average to $m(2^k-1)+1$. Let $y=m(2^k-1)+1$. We can write the two divisors as $y-x$ and $y+x$ for some $0<x<y$. Thus,
\begin{align*}
y-x&\bigm|2^ky-1,\\
y+x&\bigm|2^ky-1,
\end{align*}
since $2^ky-1=(2^k-1)(2^km+1)$. Manipulating these divisibility relations shows that
\begin{align*}
y-x&\bigm|2^kx-1,\\
y+x&\bigm|2^kx+1,
\end{align*}
where $\gcd(2^kx-1,2^kx+1)=1$. Then $\gcd(y-x,y+x)=1$ so $y^2-x^2\bigm|2^ky-1$. We clearly have $2^k-1\bigm|y-1$. By $(1)$, $x=1$ and $y=2^k$. Then $m=1$ and the two positive divisors were $2^k-1$ and $2^k+1$.
 [/INST] Too long to comment:
It is necessary that 
$$y=\left\lfloor 1+\frac{x-1}{2^k-1}\right\rfloor(2^k-1)+1$$
Proof : 
We can write
$$y-1=m(2^k-1)\tag1$$
where $m$ is a positive integer.
Also, $$y^2-x^2\mid 2^ky-1$$
implies $$2^ky-1-(y^2-x^2)\ge 0\tag2$$
From $(1)(2)$, we get
$$2^k(m2^k-m+1)-1-(m2^k-m+1)^2+x^2\ge 0,$$
i.e.
$$(2^k-1)^2m^2-2(2^k-1)(2^{k-1}-1)m-(2^k-2+x^2)\color{red}{\le} 0,$$
i.e.
$$\small\frac{2^{k-1}-1-\sqrt{(2^{k-1}-1)^2+2^k-2+x^2}}{2^k-1}\le m\le \frac{2^{k-1}-1+\sqrt{(2^{k-1}-1)^2+2^k-2+x^2}}{2^k-1}\tag3$$
Since we have
$$\frac{2^{k-1}-1+\sqrt{(2^{k-1}-1)^2+2^k-2+x^2}}{2^k-1}\le \frac{2^{k-1}-1+(2^{k-1}-1+x)}{2^k-1}\tag4$$
and
$$x\lt y=m2^k-m+1\implies \frac{x-1}{2^k-1}\lt m\tag5$$
it follows from $(3)(4)(5)$ that
$$\frac{x-1}{2^k-1}\lt m\le 1+\frac{x-1}{2^k-1}$$
from which
$$m=\left\lfloor 1+\frac{x-1}{2^k-1}\right\rfloor$$
follows.$\quad\blacksquare$
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