6. Find all positive integers $n$, that satisfy

$$
n=2^{2 x-1}-5 x-3=\left(2^{x-1}-1\right)\left(2^{x}+1\right)
$$

for some positive integer $x$.
Solution 1. By performing the product of the right-hand side and rearranging terms, the equality between the second and third members can be written as

$$
2^{x-1}=5 x+2
$$

It is easily verified that neither $x=1$ nor $x=2$ are solutions, while if $x \geq 3$, then $5 x+2$ must be a multiple of 4, so $x$ is even but not a multiple of 4, that is, $x \geq 6$. For $x=6$, it is verified that $2^{x-1}=5 x+2=32$, which would be a possible solution. Note now that if for $x \geq 6$ it is true that $2^{x-1}>5 x$ (which is true for $x=6$), then $2^{x}>10 x>5(x+1)+2$, that is, by induction $2^{x-1}$ will always be greater than $5 x+2$ for all $x \geq 7$. Therefore, the only value that the positive integer $x$ can take is 6, which in turn results in

$$
n=2^{11}-30-3=2048-33=2015
$$