Being $m$ any real number, which numbers of $n$ agrees with $\frac{x^m}{m}=\frac{x^n}{n}$? Are there infinite numbers of $n$ that agrees with $\frac{x^m}{m}=\frac{x^n}{n}$?
And yes, $x$ is a variable that can't be changed.
 A: In what follows I will assume $x\ge0$ (otherwise $x^m$ may not be defined) and use yor first formulation. I find you question ambiguous.
If you are asking whether given $m\in\mathbb{R}$ there exists $n\in\mathbb{R}$, $n\ne m$, such that
$$
m\,x^n=n\,x^m\tag1
$$
for all $x\ge0$, then the answer is no. To see it, let $x=1$ in (1) to get $n=m$.
If you are asking whether given $m\in\mathbb{R}$ there exists $n\in\mathbb{R}$, $n\ne m$, such that (1) holds for some $x\ge0$, then the answer is yes. If $m\ne0$, then for any $n\in\mathbb{R}$ such that $n/m>0$, there are two solutions of (1):
$$
x=0,\quad x=\Bigl(\frac{n}{m}\Bigr)^{\tfrac{1}{n-m}}.
$$
If $m=0$, for any $n\in\mathbb{R}$, $n\ne0$, the only solution of (1) is $x=0$
A: Looking just at positive real values of $x$, $m$ and $n$:
For $0 \lt x \le 1$, $f(x) = \frac{x^n}{n}$  is a strictly decreasing function of $n$, so there is only one solution for $n$ of $\frac{x^m}{m}=\frac{x^n}{n}$, namely $n=m$.
For $1 \lt x$, $f(x) = \frac{x^n}{n}$ is a strictly decreasing function of $n$ until $n=1/\log_e(x)$ and then a strictly increasing function thereafter, with infinite limits at $0+$ and $\infty$.  So if $m=1/\log_e(x)$ then there is only one solution for $n$ of $\frac{x^m}{m}=\frac{x^n}{n}$, namely $n=m$; otherwise there are two solutions, one of which is $n=m$.   
If you are looking at positive integer values of $x$, $m$ and $n$, the only non-trivial example is : $\frac{2^1}{1} = \frac{2^2}{2}.$
