If $n=1$, then $x=\dfrac 12$ works. Let $n>1$. Let $x=\dfrac ab$ where $a,b\in \mathbb{Z}; b>1$ and $gcd(a,b)=1$. $x^n+(x+1)^n\in \mathbb{Z}\Leftrightarrow \left(\dfrac ab\right)^n+\left(\dfrac {a+b}b\right)^n\in \mathbb{Z}\Leftrightarrow \dfrac {a^n+(a+b)^n}{b^n}\in \mathbb{Z}\Leftrightarrow b^n|a^n+(a+b)^n$. $b|a^n+(a+b)^n\Rightarrow b|a^n+a^n=2a^n\Rightarrow b|2\Rightarrow b=2$. Then $2\nmid a$. If $n$ is even, then $2^n|a^n+(a+2)^n\Rightarrow a^n+(a+2)^n\equiv 0\pmod{4}$. But $a^n\equiv (a+2)^n\equiv 1\pmod{4}$. Contradiction. Hence, $n$ must be odd. See that if $n$ is odd, then $a=2^{n-1}-1, b=2$ works $\left[2^n|(2^{n-1}-1)^n+(2^{n-1}+1)^n=2^n\cdot (...)\right]$.

Notice that when n is odd $x=-1/2$ works, since $x$ doesn’t have to be positive.