We first consider
( $\boxtimes$) Let $n \geqslant 3$ be an odd natural number. Furthermore, let $a, b$, and $c$ be three integers such that

$$a^{n-1}+a^{n-2} b+\ldots+b^{n-1}=2 c^{n-1}$$

Then $a=b=c=0$.
If this were false for some $n$, we could choose a counterexample $(a, b, c)$ for which the expression $|a|+|b|+|c|$ takes its smallest positive value. If both $a$ and $b$ were odd, the left side would be the sum of an odd number of odd summands and thus odd, while the right side is even. Therefore, at least one of the numbers $a$ and $b$ must be even. If, for example, $a$ is even, the left side has the same parity as $b^{n-1}$, and the right side is even, so $b$ must also be even. Similarly, one sees that $a$ must be even if $b$ is assumed to be even. Therefore, both $a$ and $b$ are even. Consequently, the left side is divisible by $2^{n-1}$ and thus, in particular, by 4, which implies that $c^{n-1}$ and hence $c$ is also even. If we now introduce three integers $a^{\prime}, b^{\prime}$, and $c^{\prime}$ with $a=2 a^{\prime}, b=2 b^{\prime}$, and $c=2 c^{\prime}$, then first

$$a^{\prime n-1}+a^{\prime n-2} b^{\prime}+\ldots+b^{\prime n-1}=2 c^{\prime n-1}$$

and secondly

$$0<\left|a^{\prime}\right|+\left|b^{\prime}\right|+\left|c^{\prime}\right|<|a|+|b|+|c| .$$

The triple $\left(a^{\prime}, b^{\prime}, c^{\prime}\right)$ thus contradicts the choice of $(a, b, c)$, and $(\boxtimes)$ is proved.
We now turn to the actual problem. In the case $n=1$, the claim is trivial, and from now on we focus only on the case $n \geqslant 3$. Let $x$ and $y$ be two rational numbers such that $x^{n}+2 x=y^{n}+2 y$. By bringing everything to one side and then factoring, we obtain

$$(x-y)\left(x^{n-1}+x^{n-2} y+\ldots+y^{n-1}-2\right)=0$$

If $x$ and $y$ were different, which we will assume from now on, then

$$x^{n-1}+x^{n-2} y+\ldots+y^{n-1}=2$$

must hold. Since any two rational numbers can be brought to a common denominator, there exist three integers $a, b$, and $c$ with $c \neq 0, x=a / c$, and $y=b / c$. Substituting these fractions into (1) and then multiplying by $c^{n-1}$, we get

$$a^{n-1}+a^{n-2} b+\ldots+b^{n-1}=2 c^{n-1}$$

From $(\boxtimes)$ it follows, in particular, that $c=0$. This contradiction shows that $x$ and $y$ must indeed be equal, thus solving the problem.