First we prove: $x + y > 1$.
Note that,
$$(x + y)(x^n - y^n) = x^{n+1} - y^{n+1} + xy(x^{n-1} - y^{n-1}).$$
$$\text{Then } x + y = \frac{x^{n+1} - y^{n+1}}{x^n - y^n} + xy \times \frac{x^{n-1} - y^{n-1}}{x^n - y^n} = 1 + xy \times \frac{x^{n-1} - y^{n-1}}{x^n - y^n}.$$
Since $n > 1$, we know that $x^n - y^n$ and $x^{n-1} - y^{n-1}$ have the same sign.
Thus,
$$\frac{x^{n-1} - y^{n-1}}{x^n - y^n} > 0.$$
Therefore $x + y > 1$.

Next we prove:
$$x + y < \frac{2n}{n + 1}.$$
First we prove a lemma.
Lemma: Let $x, y > 0, n$ be a positive integer. Then
$$\frac{x^{n+1} - y^{n+1}}{n + 1} > \left(\frac{x^n - y^n}{n}\right) \left(\frac{x + y}{2}\right). \quad (1)$$

Proof:
$$\begin{align*} & (1) \text{ is equivalent to} \quad \frac{x^n + x^{n-1}y + x^{n-2}y^2 + \cdots + y^n}{n+1} \\ & \quad > \left( \frac{x^{n-1} + x^{n-2}y^2 + \cdots + xy^{n-2} + y^{n-1}}{n} \right) \left( \frac{x+y}{2} \right) \\ & \quad \Leftrightarrow \frac{x^n + x^{n-1}y + x^{n-2}y^2 + \cdots + y^n}{n+1} \\ & \quad > \frac{2(x^n + x^{n-1}y + x^{n-2}y^2 + \cdots + y^n) - (x^n + y^n)}{2n} \\ & \quad \Leftrightarrow \frac{x^n + x^{n-1}y + x^{n-2}y^2 + \cdots + y^n}{n+1} < \frac{x^n + y^n}{2} \\ & \quad \Leftrightarrow 2(x^n + x^{n-1}y + x^{n-2}y^2 + \cdots + y^n) \\ & \quad < (n+1)(x^n + y^n). \tag{2} \end{align*}$$
$$\begin{align*} & \text{From } (x^i - y^i)(x^{n-i} - y^{n-i}) > 0 \\ & \Rightarrow x^i y^{n-i} + x^{n-i} y^i < x^n + y^n. \end{align*}$$
Therefore (2) holds, and the lemma is proved.

Return to the original problem.
Without loss of generality, assume $x > y > 0$. Then
$$x^n - y^n > 0, x^{n+1} - y^{n+1} > 0.$$
From the lemma and the given condition, we get
$$\begin{gather*} \frac{1}{n+1} > \frac{x+y}{2n} \\ \Rightarrow x+y < \frac{2n}{n+1}. \end{gather*}$$
Therefore $1 < x + y < \frac{2n}{n + 1}$.