I'm trying to prove (or disprove) the following "conjecture".

Given the following set of powers of two:

$$A = \{ x \mid x = 2^n \text{ and } 2^{n-1} < 3^m < x < 3^{m+1} < 2^{n+1}\}$$

(informally $2^n$ is "snapped" in an interval $[3^m,3^{m+1})$ that contains only one power of two)

**Q1:** Given an integer $k \geq 1$, is there a way to pick an infinite subset of $B \subseteq A$ (or possibly the whole $A$) such that the following holds:

For all integers $x,y,p,q \geq 1$, with $x,y \in B$, $x \neq y$ and all the factors of $p, q$ are smaller than $k$, does there always exist $n \geq 1$ such that:

$$x + np \in A,\quad y + nq \notin A\quad \text{?}$$

Informally whatever pair of elements of $B$ and deltas $p,q$ I choose, the two arithmetic progressions $x' = x + np$ and $y' = y +nq$ don't guarantee that $x', y'$ are always both inside or both outside of $A$.

**Q2:** And is the "conjecture" easier if the set is:
$$A' = \{ x \mid 2^{n-1} < 3^m \leq x < 2^n < 3^{m+1} < 2^{n+1}$$

The only fuzzy idea for Q2 I have is that if $x,y$ are big enough and belong to the same $3^m \leq x<y < 2^n$ interval, then $p,q$ (that have factors smaller than $k$) don't give enough "resolution" (??) to keep $x'$ and $y'$ "aligned".

Sorry for the possibly inaccurate terminology but I'm not an expert.