Let $y=\frac{p}{q}$ and $x=\frac{r}{s}$ where $\gcd(p,q)=\gcd(r,s)=1$. We have $$p\cdot \left(\frac{r}{s}\right)^{\frac{p}{q}}=p+q \Longleftrightarrow p^q \cdot r^p=(p+q)^q \cdot s^p$$And since $\gcd(p+q,p)=\gcd(r,s)=1$ we must have $p^q=s^p$ and $r^p=(p+q)^q$. Lemma: If $x^a=y^b$ for some $x,y,a,b$ naturals , then there exists a natural $z$ such that $x=z^m$ and $y=z^n$ where $m=\frac{b}{\gcd(a,b)}$ and $n=\frac{a}{\gcd(a,b)}$. Proof: $\frac{a}{b}=\frac{n}{m}$ where $\gcd(m,n)=1$ , from $x^a=y^b$ we have $x^n=y^m$ and considering the prime factorisation of $x,y$ we have $n\cdot \alpha_i=m\cdot\beta_i$. Take $r_i=\frac{\alpha_i}{m}=\frac{\beta_i}{n}$, and taking $z=\prod {p_i}^{r_i}$ , we have $x=z^m$ and $y=z^n$.$\blacksquare$ Using above lemma in $p^q=s^p$ , there exists a $z$ such that $p=z^p$ , if $z \neq 1$ we have $p=z^p=z^{z^p}$ and continuing like this , $p$ is unbounded ,contradiction. So , we must have $z=1$ wich means $p=s=1$ and from $r^p=(p+q)^q$ we have $r=(q+1)^q$ . So, the only solutions are $x=(q+1)^q$ and $y=\frac{1}{q}$, where $q$ is any positive integer.