Hellenic Mathematical Olympiad

Let $y = \frac{p}{q}$, with $p, q \in \mathbb{N}^*$, $(p, q) = 1$. Then the equation is written in the form:
$$x^q = \frac{y+1}{y} = \frac{\frac{p}{q}+1}{\frac{p}{q}} = \frac{p+q}{p} \Rightarrow x = \sqrt[q]{\frac{(p+q)^q}{p^q}}.$$
Therefore the right hand side must be rational. Since $(p+q, p)=1$, it follows that the numerator and the denominator of the fraction must be powers of $p$. Hence
$$p+q = a^p \quad \text{and} \quad p = b^p,\ a, b \in \mathbb{N}^*,\ a > 1.$$
If $b \ge 2$, then $b^p \ge 2^p > p$ (absurd), and hence $b=1$, which implies $p=1$ and $q = a - 1$. Hence the solutions of the given equation are given parametrically in the form: $(x, y) = \left( a^{a-1}, \frac{1}{a-1} \right),\ a \in \mathbb{N}^*,\ a > 1$.