Let $x$ be a real number such that $x^n$ and $(x+1)^n$ are rational. Without loss of generality, I assume that $x \neq 0$ and $n>1$.

Let $F = \mathbb Q(x, \zeta)$ be the subfield of $\mathbb C$ generated by $x$ and a primitive $n$-th root of unity $\zeta$.

It is not difficult to see that $F/\mathbb Q$ is Galois; indeed, $F$ is the splitting field of the polynomial $$X^n - x^n \in \mathbb Q[X].$$ The conjugates of $x$ in $F$ are all of the form $\zeta^a x$ for some powers $\zeta^a$ of $\zeta$. Indeed, if $x'$ is another root of $X^n - x^n,$ in $F$, then

$$(x/x')^n = x^n/x^n = 1$$

so that $x/x'$ is an $n$-th root of unity, which is necessarily of the form $\zeta^a$ for some $a \in \mathbb Z/n\mathbb Z$.

Assume now that $x$ is **not** rational. Then, since $F$ is Galois, there exists an automorphism $\sigma$ of $F$ such that $x^\sigma \neq x$. Choose any such automorphism. By the above remark, we can write $x^\sigma = \zeta^a x$, for some $a \not \equiv 0 \pmod n$.

Since also $(1+x)^\sigma = 1+ x^\sigma \neq 1 + x$, and since $(1+x)^n$ is also rational, the same argument applied to $1+x$ shows that $(1+x)^\sigma = \zeta^b (1+x)$ for some $b \not \equiv 0 \pmod n$. And since $\sigma$ is a field automorphism it follows that

$$\zeta^b (1+x) = (1+x)^\sigma = 1 + x^\sigma = 1 + \zeta^a x$$

and therefore

$$x(\zeta^b - \zeta^a) = 1 - \zeta^b.$$

But $\zeta^b \neq 1$ because $b \not \equiv 0\pmod n$, so we may divide by $1-\zeta^b$ to get

$$x^{-1} = \frac{\zeta^b - \zeta^a}{1-\zeta^b}.$$

Now, since $x$ is real, this complex number is invariant under complex conjugation, hence

$$x^{-1} = \frac{\zeta^b - \zeta^a}{1-\zeta^b} = \frac{\zeta^{-b} - \zeta^{-a}}{1-\zeta^{-b}} = \frac{1 - \zeta^{b-a}}{\zeta^b-1} = \zeta^{-a}\frac{\zeta^a - \zeta^b}{\zeta^b-1} = \zeta^{-a} x^{-1}.$$

But this implies that $\zeta^a = 1$, which contradicts that $x^\sigma \neq x$. So we are done. $\qquad \blacksquare$

The following stronger statement actually follows from the proof:

*For each $n$, there are finitely many non-rational complex numbers $x$ such that $x^n$ and $(x+1)^n$ are rational. These complex numbers belong to the cyclotomic field $\mathbb Q(\zeta_n)$, and none of them are real.*

Indeed, there are finitely many choices for $a$ and $b$.

In this related answer, Tenuous Puffin proves that there exist only $26$ real numbers having this property, allowing for any value of $n$.