When I look at the encyclopedia of integers, it is clear that this grows even though it doesn't grow continuously.
It seems to me that it should be possible to establish a lower bound based on identifying different cases where $m$ is the minimum power of $2$ greater than $3^n$:
Case 1: $2^m > 3^n$ and $2^{m+1} > 3^{n+1} > 2^m$
In this case, $2^m > \dfrac{3}{2}3^n$ so $2^m - 3^n > \dfrac{3^n}{2}$
Case 2: $2^m > 3^n$, $2^{m+2} > 3^{n+1} > 2^{m+1}$ and $2^{m+3} > 3^{n+2} > 2^{m+2}$
In this case, $2^m > \dfrac{9}{8}3^n$ so $2^m - 3^n > \dfrac{3^n}{8}$
I am able to define the remaining case but I am not able to find a pattern.
Case 3: $2^m > 3^n$, $2^{m+2} > 3^{n+1} > 2^{m+1}$, and $2^{m+4} > 3^{n+2} > 2^{m+3}$
In this case, I am not able to find a generalizeable lower bound. I'll give an example. Case $3$ holds for $m=8, n=5$. In this case, eventually, I find that $2^{m} > \dfrac{3^{12}}{2^{19}}3^n = \dfrac{531,441}{524,288}3^n$
When I check the next example of case 3 at $m=16, n=10$, I get $2^{m} > \dfrac{3^7}{2^{11}}3^n = \dfrac{2187}{2048}3^n$
Is there a straight forward way to find a lower bound for Case 3? Is this an open problem?
With $m=\lceil n \log_23\rceil$, bounds for $2^{\lceil n \log_23\rceil}-3^n$ can be found here: Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$