<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] This is just a hint, which may be useful, not a full answer.
$$(2^k-1)(2^k+1)=(2^k-x)(2^k+x)t$$
From this we see that $x$ should be odd. And easy to prove that the $$(2^k-x,2^k+x)=1 $$
As the $2^k+x > 2^k-1$ and $2^k+x > 2^k+1$ for the $x>1$ then $$(2^k+x,2^k-1)=a > 1$$ and $$(2^k+x,2^k+1)=b > 1$$ and $(a,b)=1$ and $ab=2^k+x$.
Let assume that $$(2^k-1,2^k-x)=c$$ and $$(2^k+1,2^k-x)=d$$ then obviously $(a,c)=1$, $(a,d)=1$, $(b,c)=1$, $(b,d)=1$, $(d,c)=1$ and $cd=2^k-x$.
And $a,b,c,d$ are odd.
$$ac | 2^k-1$$
$$bd | 2^k+1$$
$$ab+cd=2^{k+1}$$
$$ab-cd=2x$$
From this it seems that there should be some solution, but probably for the $k$-s for which the $2^k-1$ and $2^k+1$ has enough divisors.
In this scope it is also interesting to consider the Bang's theorem
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