Let $x>1$ be a real number. For a (fixed) natural number $n>1$, is it possible to fully or partially classify all solutions to the equation $$\left\{x^n\right\}=\{x\}^n$$ where $\{\cdot\}$ represents the fractional part function, i.e., $\{x\}=x-\lfloor x\rfloor$ where $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.
Of course, all integers are solutions. For the case $n=2$, another family of solutions is given by $$x_m=\left(m+\frac 1m\right)$$ for any $m\in \mathbb N$. This is because $$x_m^2 = m^2 + 2 + \frac 1{m^2}$$ and hence $\left\{x_m^2\right\}=1/m^2$. But, are these all? What about the general case? Even partial solutions are welcome.
Let $n\in\Bbb{N}$ and suppose $x\in\Bbb{R}$ is such that $\{x^n\}=\{x\}^n$. Let $m:=\lfloor x\rfloor$ and $y:=\{x\}$ so that $m$ is an integer and $y\in[0,1)$, and $$(m+y)^n-y^n=k,$$ for some integer $k$. Then $y$ is a root of the polynomial $$f:=(m+X)^n-X^n-k=-k+\sum_{i=0}^{n-1}\binom{n}{i}m^{n-i}X^i.$$ Because $x>1$ we also have $m\geq1$, and so as a real function, this polynomial is strictly increasing on the interval $[0,1)$. Moreover $$f(0)=m^n-k\qquad\text{ and }\qquad f(1)=(m+1)^n-1-k,$$ so if $f$ has a root in $[0,1)$ then $k\in[m^n,(m+1)^n-1)$.
Conversely, given $n\in\Bbb{N}$ and $m\in\Bbb{Z}$ with $m\geq1$, if $k\in[m^n,(m+1)^n-1)$ then the polynomial $$f=(m+X)^n-X^n-k=-k+\sum_{i=0}^{n-1}\binom{n}{i}m^{n-i}X^i,$$ has a unique root $y\in[0,1)$ and $x:=m+y$ satisfies $\{x^n\}=\{x\}^n$.
In particular, for $n=2$ you can take any integer $m\geq1$ and any integer $k\in[m^2,m^2+2m)$ and take the unique root $y_k\in[0,1)$ of $$-k+m^2+2mX=0,$$ which is of course $y_k:=\frac{k-m^2}{2m}$. For $k=m^2+2$ you get $y=\frac1m$ yielding your solution $x=m+\frac1m$.