Suppose $\,r\geq a\,\,,\,\,a,b,r,s\in\mathbb N\,$:
$2^r(2s-1)-1=2^a(2b-1)-1\Longrightarrow 2^{r-a}(2s-1)=2b-1\Longrightarrow r=a$as the RHS is odd and so must be the LHS $\,\Longrightarrow 2s-1=2b-1\Longrightarrow b=s\,$ , and uniqueness has been proved , so that $\,X_r\cap X_a=\emptyset\,\,\,,\,\,\forall\,r\neq a\,$
Also, if $\,t\,$ is an odd natural, then $\,t=2k-1\,$ , for some natural $\,k\,$ . Let $\,r-1\in\mathbb N\,$ be the maximal power of $\,2\,$ that divides $\,k\,$ , then $t=2(2^{r-1}m)-1$and since $\,m\,$ is odd, $\,m=2s-1\,\,,\,\,s\in\mathbb N\,$ , so $t=2\left(2^{r-1}(2s-1)\right)-1=2^r(2s-1)-1\in X_r$