If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$

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.$\quad\blacksquare$