<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] For $k\ge 1$ and $0<x<2^k$, suppose$$(2^{2k}-x^2) |  (2^{2k}-1)$$for some $x>1$, and hence that $2^{2k}-x^2$ is composed only of the prime factors of $2^{2k}-1$ but lacks one or more of them (or perhaps contains all of the distinct factors but with one or more of them to a lesser power).
Thus let$$2^{2k}-1=(2^k-1)(2^k+1)=pqr\cdot stu$$and suppose, for some $x>1$, that$$2^{2k}-x^2=pq\cdot stu$$and hence divides $2^{2k}-1$.
Then since for $x\ge1$,$$2^{2k}-x^2=(2^k-x)(2^k+x)$$and hence the sum$$(2^k-x)+(2^k+x)=2^{k+1}$$then$$pqr+stu=pq+stu=2^{k+1}$$which is impossible.
Hence it is clear that removing one or more prime factors from either $2^k-1$ or $2^k+1$, while leaving the other addend intact, must make their sum less than $2^{k+1}$. Still less can their sum be $2^{k+1}$ if one or more prime factors are removed from both $2^k-1$ and $2^k+1$.
The remaining possibility is that removing one or more prime factors from both $2^k-1$ and $2^k+1$ and re-arranging the remaining prime factors, might yield a sum $=2^{k+1}$.
E.g. for $k=6$, $(2^k-1)(2^k+1)=63\cdot65=3^2\cdot7\cdot5\cdot13$, and$$3^2\cdot7+5\cdot13=2^{k+1}=2^7$$Removing one $3$-factor and re-arranging the four remaining distinct prime factors in the seven possible ways we get$$3\cdot7+5\cdot13=86$$$$3\cdot5+7\cdot13=106$$$$3\cdot13+5\cdot13=74$$$$3+7\cdot5\cdot13=458$$$$5+3\cdot7\cdot13=278$$$$7+3\cdot5\cdot13=202$$$$13+3\cdot5\cdot7=118$$Noteworthy here is that all sums are odd multiples of $2^1$.
For $k=10$, removing one of the $5$-factors from $2^{2k}-1=1023\cdot1025=3\cdot11\cdot31\cdot5^2\cdot41$, I find the fifteen possible sums of two addends containing the remaining five distinct primes are all odd multiples of $2^2$.
If it could be shown, then, that all such sums dividing $2^{2k}-1$ are odd multiples of some power of $2$, and hence not equal to $2^{k+1}$, it would follow that, for $k\ge1$ and $x>1$, $2^{2k}-x^2$ does not divide $2^{2k}-1$.
This of course is not a full answer, but it lays out a possible approach.
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