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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 almost yes. Given $\varepsilon > 0$, 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 ratio $\alpha = \frac{1 + \sqrt{5}}{2}$, the sequence $(\alpha^n)_n$ is not equidistributed modulo $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates for the smallest $m(\alpha, \varepsilon)$ and $n(\alpha, \varepsilon)$ are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (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.

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 almost yes. Given $\varepsilon > 0$, 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 equidistributed modulo $1$ in the sense of H. Weyl for almost every $\alpha > 1$, 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 ratio $\alpha = \frac{1 + \sqrt{5}}{2}$, the sequence $(\alpha^n)_n$ is not equidistributed modulo $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates for the smallest $m(\alpha, \varepsilon)$ and $n(\alpha, \varepsilon)$ are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (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.

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 almost yes. Given $\varepsilon > 0$, 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 $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (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.

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 almost yes. Given $\varepsilon > 0$, 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 ratio $\alpha = \frac{1 + \sqrt{5}}{2}$, the sequence $(\alpha^n)_n$ is not equidistributed modulo $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates for the smallest $m(\alpha, \varepsilon)$ and $n(\alpha, \varepsilon)$ are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (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.

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 $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (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.

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 almost yes. Given $\varepsilon > 0$, 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 $1$, it is not even dense, since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts do accumulate on $0$. A similar line of reasoning applies to Pisot-Vijayaraghavan numbers; this is Robert Israel's comment. Indeed, when $\alpha$ is a Pisot-Vijayaraghavan number, say of of degree $d$ with conjugates $\alpha_1 = \alpha, \alpha_2, \dots,\alpha_d$, estimates are easy to come by since $\sum_{i = 1}^d \alpha_i^k$ is an integer for all $k$ while $\sum_{i > 1}^d \alpha_i^k$ tends exponentially fast towards $0$. (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.