Is there any integer $x$ such that $2^n$ divides $3^n(x+1)$ for all integers $n$?

Question

I am wondering whether: $$\exists x \in \mathbb{N}^* / \, \forall n \in \mathbb{N}^*,\, 2^n\mid 3^n(x+1)$$

I re-wrote it as $$2^n \leq 3^n(x+1)$$ but it doesn't seem like a good approach. Any ideas?

Answer 1

Since $ 2^n $ is relative prime with $ 3^n $ for each $ n $, necessarily $ 2^n $ divides $ x + 1 $ for each $ n $. Then $ x + 1 $ has infinite different divisors, so $ x + 1 = 0 $, that is $ x = -1 $