Solving A Certain Diophantine Equation I am stuck on finding the solution of the Diophantine equation:
$d(2^{k+1}-1)-b^2(2^{k+1}-2)=1$.
where $k\geq 1$ and $b^2>d$ for $b$ an odd composite integer. 
Is there a solution to this Diophantine equation? If so, what are they and how can I find it? 
Thank you so much.
 A: (Old answer revised with new information.)
Given,
$$d(2^{k+1}-1)-b^2(2^{k+1}-2)=1$$
There are solutions for all $k$. It is best to simplify with $k=p-1$ to get $\color{brown}{M_p = 2^p-1}$. Hence,
$$d(M_p)-b^2(M_p-1)=1\tag1$$

I. Family 1: The two general solns in the old answer can be combined into one. For any $p$, there is an infinite number of positive integer $b,d$ that solves $(1)$. For integer $n$. 
$$b = M_p n \pm 1$$
$$d = (M_p)(M_p-1)n^2\pm2(M_p-1)n+1$$
II. Family 2: But there are additional parametric solutions for even $p$,
$$b = M_p n \pm  2\,^{p/2}$$
$$d = (M_p)(M_p-1)n^2\pm 2\cdot2\,^{p/2}(M_p-1)n+M_p$$

One may ask, "Using integer $n$, does this give all the solutions?". If $M_p$ is composite, the answer is no. For example, for $p=23$, the smallest non-family solution is,
$$b = M_{23}n+1 = 3034178,\quad\text{where}\;n=17/47 $$
However, if $M_p>3$ is a Mersenne prime, for $p = 3,5,7,13,17,19,\dots$ then it seems $b = M_p n \pm 1$ is the complete solution. (I tested $b<10^6$ and there were no non-family solutions for those $p$.)
P.S. These families were found using Mathematica and some intuitive guessing.
