The following seems to be implied by most of the direct comments to OP's question, but I prefer to voice it, loud and clear:
The answer is yes, we can find such integers $m = m(\alpha, \varepsilon)$ and $n = n(\alpha, \varepsilon)> 0$, for almost every real number $\alpha > 1$, in the sense of Lebesgue measure.
This is so, because the sequence $(\alpha^n)_n$ is then equidistributed modulo $1$ in the sense of H. Weyl, see "Uniform distribution of sequences" by L. Kuipers et H. Niederreiter (Wiley, 1974).
Equidistribution modulo $1$ implies density of the set $\{\alpha^n - \lfloor \alpha^n \rfloor : n \in \mathbb{N}\} \subset \lbrack 0, 1\rbrack$, which in turn implies that the sequence $(\alpha^n - \lfloor\alpha^n\rfloor)_n$ of fractional parts accumulates on $0$.
As already observed above, for the golden number $\alpha = \frac{1 + \sqrt{5}}{2}$, the sequence $(\alpha^n)_n$ is not equidistributed modulo $1$ since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers (this is Robert Israel's comment, see also Noam D. Elkies'comment for Salem numbers).
It is not known whether $((\frac{3}{2})^n)_n$ is equidistributed modulo $1$ (See Gerry Myerson's comment for references). It is also not known whether $(e^n)_n$ or $(\pi^n)_n$ is equidistributed modulo $1$. Actually, no explicit example $\alpha > 1$ is known to be equidistributed modulo $1$. (All my remarks are borrowed from a lecture on ergodic theory by Pierre de La Harpe.)