If for every k the interval $[a,ak]$ contains $n$ specials numbers how many special numbers $[az,akz]$ must contain? The purpose of my question is to determine if a specific kind of reasoning is true or false. 
Let's say that for every positive natural number $a$, there is a at least $n$ "special numbers" in the interval of natural numbers $[a,ak]$ where $k$ is a constant. 
Is it true that, by multiplying the boundaries by $z$, the number of "special numbers" in the interval $[az, akz]$ is also multiplied by $z$?
If no, why ? And if yes, what are the conditions on $z$ for this being true ?

Let's define the set of special numbers as $\{1, 2, 4, 8, 16, \dots\}$ and take $k = 2$.
For $a = 1$, the interval $[a,ak] = [1,2]$ contains two special numbers, for $a = 3$, the interval $[a, ak] = [3, 6]$ contain one special number. 
Take $a = 1$. If the answer to you question was YES, then, for all $z\in\mathbb{N}$, the intervals of form $[az, azk] = [z, 2z]$ should contain at least $2$ special numbers, which is not the case.

Maybe, we should prove that all intervals $[a, 2a]$ contain at least one special number, otherwise the upper set is not a counterexample.
If there is a number $a\in \mathbb{N}$ such that $[a, 2a]$ does not contain a power of two, that it follows that $$2^m < a\quad \text{and} \quad 2a < 2^{m + 1}$$
for all $m\in\mathbb{N}$, which obviously leads to contradiction if we divide the second inequality by $2$.
