I found a pattern that I want to prove:
$$f(x) = 2^{\lceil \log_2(3^x)\rceil} - 3^x\quad \{x\in\mathbb{Z}^+\} $$ $$ \lim_{x\rightarrow\infty} f(x) = \infty $$
Discussion:
$$ f(x) = 2^{\lceil \log_2(3^x)\rceil} - 3^x = 3^x(2^{\lceil \log_2(3^x)\rceil-\log_2(3^x)}-1)$$
$$0<\lceil \log_2(3^x)\rceil-\log_2(3^x)<1\ \Rightarrow\ 0<f(x)<3^x$$
$f(x)$'s lower bound is zero and its upper bound tends to infinity. As far as I know, $f(x)$ can oscillate anywhere in between. However, after checking a few first thousands values of $f$, I am convinced that the function indeed tends to infinity. I guess the fact that $x$ is an integer plays a role in that.
If $\lceil \log_2(3^x) \rceil$ was really close to $\log_2(3^x)$, $f(x)$ would be really close to $0$, so the limit would not hold. Experimental results suggest the limit exists, so I guess there are some restrictions on how close $\log_2(3^x)$ can be to its ceiling integer.
I don't know how to proceed from that.