Here's what I've got so far:
Let $r = \frac{a}{b}$, where $a$ and $b$ are integers. We then have $$r^{\frac{1}{r-1}} = \frac{a^{\frac{b}{a-b}}}{b^{\frac{b}{a-b}}}$$
Clearly, $a-b=1$ and $a-b=-1$ work. But it might be the case that $a-b \neq 1$ or $a-b \neq -1$, but $\frac{a^{\frac{b}{a-b}}}{b^{\frac{b}{a-b}}}$ still be rational, if it turns out that $a=p^{\frac{1}{a-b}}$ and $b=q^{\frac{1}{a-b}}$.
But how do I obtain a contradiction if $a-b > 1$, or $a-b<-1$? I let $a-b=n$, which means that $a=p^{n}$ and $b=q^{n}$, so $$a-b = p^n - q^n = n$$
This is strangely curious, because it's worked out well so far. I can see that the expression in the middle can be factored, but how will that help with the contradiction?
If you have your own way of approaching this problem, feel free to comment. Thanks.
It is much easier to start with $\frac{1}{1-r}=\frac{a}{b}$ so $r=\frac{a-b}{a}$. then you want $a-b$ and $a$ to be perfect $b$th power, or $b-a$ and $-a$ to be perfect $b$th powers. But $\left|(k+1)^b-k^b\right|$ is greater than $b$ when $b>1$ and $k\neq -1,0$, and the $k=-1,0$ cases are no good when $b>1$. So if $b>1$ then there are no solutions.
The case when $b=1$, however, gives a solution for any $a\neq 0$.