<s>[INST] If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?
 [/INST] The equation resolves to $(2^k+x)(2^k-x) \mid (2^k+1)(2^k-1)$.  Since we have the first factor larger than the rest, we would look for a common factor in $(x- 1)$ and $(2^k-1)$ or $(x+1)$ and $2^k+1$,
We could suppose that the first is a product of say $(ab)(cd)$, and that the second is a product of $(ac)(bd)e$, where $ac$ divides $2^k+1$ and $bd$ divides $2^k-1$.  The common divisor between the first two and the last two factors, is $x+1$, and between the second and third, and the first and fourth, $x-1$.
But this common factor must also divide $2^k+1$ and $2^k-1$, and so must divide $2$.
So there is no other number which divides pairs of divisors (ie $2^k \pm 1$ and $2^k\pm x$).
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