Solution with Dave12345 The given equation is equivalent to $(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})=2(x-y)$. Assume $x \neq y$, then this is equivalent to $x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}=2$. Now looking $mod \ x$ and $mod \ y$ we get $x|y^{n-1}-2$ and $y|x^{n-1}-2$. Now let $x=\frac{p}{q}$ and $y=\frac{a}{b}$ such that $(p,q)=(a,b)=1$. We easily get $q=b^{n-1}k$ and similarly $b=p^{n-1}t$ $k,t \in \mathbb{N}$. After multiplying we get $1=b^{n-2}q^{n-2}kt$ so all that is left is casework.

@above the question says that $x,y$ are rational numbers.