Let $xy=z^n$ where $x$, $y$ and $z$ belong to a Dedekind domain $R$, with $n>1$, and $(x,y)=1$. We can also assume that the ideal class group of $R$ is torsion-free.
Then I’d like to show that $x=ur^n$, where $r\in R$ and $u$ is a unit in $R$. I thought of starting from the unique factorization of $(x)$ and $(y)$ into prime ideals, equating this to the prime ideal factorization of $(z)^n$. But I couldn’t get much further.
Hint. For any prime divisor $P$ of $(x)$, we have $v_P(y)=0$ (why ?). One may deduce that $(x)=I^n$ for some ideal $I$ of $R$. Now use the assumption on the ideal class group to get that $I$ is a principal ideal. If $I=(r)$, we then get $(x)=(r^n)$, so...