The value of $$\inf \left\{ |\pi^m-e^n|: m,n\in\mathbb{N} \right\}$$ is a known unsolved problem. But transcendental numbers are known to cause problems of this sort.
Is the value of $$\inf \left\{ |\sqrt{2}^m-\sqrt{3}^n|: m,n\in\mathbb{N} \right\}$$ known? Or at least is it known if $|\sqrt{2}^m-\sqrt{3}^n|$ can be arbitrarily small?
Note you have
$$d = \left|\sqrt{2}^m-\sqrt{3}^n\right| = \frac{\left|2^m - 3^n\right|}{\sqrt{2}^m + \sqrt{3}^n} \tag{1}\label{eq1A}$$
As stated near the bottom of Differences Between Powers,
Also, a closely related post is $\liminf |2^m - 3^n|$. Its accepted answer uses Baker's theorem to show that
which is very similar to what Tijdeman determined.
Since you're looking for $d$ in \eqref{eq1A} to be very small, let
$$\sqrt{3}^n = (1 + \epsilon)\sqrt{2}^m \tag{2}\label{eq2A}$$
where $\epsilon \approx 0$. Also, to get smaller values of $d$, $\epsilon$ should get closer to $0$ as $m$ increases.
From \eqref{eq1A}, using Tijdeman's result and \eqref{eq2A}, gives
$$\begin{equation}\begin{aligned} \left|\sqrt{2}^m-\sqrt{3}^n\right| & \ge \frac{2^m}{m^c(\sqrt{2}^m + \sqrt{3}^n)} \\ & = \frac{2^m}{m^c(2 + \epsilon)\left(2^{m/2}\right)} \\ & = \frac{2^{m/2-1}}{m^c\left(1 + \frac{\epsilon}{2}\right)} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
The numerator is an exponential in $m$ while, since $c$ is a fixed real number and $\epsilon$ is relatively small (and ideally decreasing), the denominator is basically a polynomial in $m$. Since exponentials grow faster than polynomials, this means \eqref{eq3A} shows the minimum difference grows without bound as $m$ increases. This also means the $\epsilon$ in \eqref{eq2A} cannot stay close to $0$ and, actually, must be increasing. Thus, this proves $\left|\sqrt{2}^m-\sqrt{3}^n\right|$ can't be made arbitrarily small.
Regarding the smallest value $d$ can be, this can be determined by checking the smallest values of $m$, with the required number to check depending on what the value of $c$ is. However, I don't know if anybody has done this and, if so, what the result is.