Edited from prior incorrect answer:

## Summary

At $m=n=2, f=\tfrac{1}{4}$, but it doesn’t seem to go below that, and in fact gets large quickly as we count up in $n$ and use the minimizing value of $m$ for that $n$.

I’m relatively confident, but not certain, that there is no such number. See below.

## Minimization Attempt

To first approach this, look for any minima in the function on a continuous domain and, if we find any, check the nearest natural numbers.

Find the critical points of $(\frac{3}{2})^n - 2^m$ and of it’s inverse. Consider any min’s of the former and maxes of the latter for any case $ \tfrac{-1}{4} < f < \tfrac{1}{4} $

$$f_n \equiv \frac{\partial f}{\partial n} = log (3/2) (3/2)^n$$ has no zero point but exists everywhere, so no critical point.

$f_m = -\log (2) 2^m$, same.

(If we had critical points, we’d check the sign of the $D=f_{xx}f_{yy} - f_{xy}^2$ at the m and n of the critical point(s) to see if min or max.)

We have no mins nor maxes in the range of interest.

## Looking for Solutions

If the domain were continuous, one solution would be to set $$(\frac{3}{2})^n = 2^m \implies f=0$$

$$n \log(3/2) = m \log2 , m/n = \log(3/2)/\log(2)$$

With our problem, pick natural numbers to approximate that.

For any $n$, the best $m$ is $Round(n \log(3/2)/\log(2))$, where $Round()$ is rounding to the nearest natural number.

So, the best attempt at any $n$ for $f(m,n) is:

$$g(n)= |(\frac{3}{2})^n - 2^{(Round(n \log(3/2)/\log(2))}|$$

This also cannot be minimized or maximized, in this case because $g, g_n$ not continuous. This at least limits the $m,n$ space to search within, but cannot find a solution analytically.

## Convergence?

One possible answer would be if this converged to some value less than 1/4. The needed convergence:

$$\lim_{n \to -\infty} |(\frac{3}{2})^n - 2^{(Round(n \log(3/2)/\log(2))}|<\tfrac{1}{4}$$

If anyone can solve that I’d be impressed? Checking some high values, it appears to get big very fast. In fact, at $m=n=2, f=0.25$, and tends strongly to increase at higher $n$ (but **not** monotonically in $n$) and appears to diverge in the limit.

But I can’t guarantee there isn’t some very large, fortuitous $n$ that surprisingly works. Other conjectures in math, such as the Polya Conjecture, have held to very high numbers ($10^{300}$ in that case), only for a surprise nearly knife-edge counterexample to be found (~1.845 E365). And there may yet be an analytical solution out there for this problem.