The unique sequence satisfying the problem is
$$\{x_1, x_2, \dots, x_{101}\} = (1, k, \dots, k) \text{ where } k = 2 + 3 + \dots + 101 = 5150.$$
Let $N$ denote the set of all positive integers. $k = 2 + 3 + \dots + 101 = 5150$. Then we obtain
$$1^n + 2k^n + \dots + 101k^n = 1 + (2+3+\dots+101)k^n = 1 + k \cdot k^n = k^{n+1} + 1$$
for any $n$, so $(1, k, \dots, k)$ satisfies the problem. Below we prove that this solution is unique.

Let $(x_1, \dots, x_{101})$ satisfy the problem. Then for any $n \in N$, there exists $y_n \in N$ such that
$$x_1^n + 2x_2^n + \cdots + 101x_{101}^n = y_n^{n+1} + 1.$$
Note that $x_1^n + 2x_2^n + \dots + 101x_{101}^n < (x_1^n + 2x_2 + \dots + 101x_{101})^{n+1}$, from which it can be deduced that the sequence $\{y_n\}$ is a bounded sequence. That is, there exists $y \in N$ such that $y_n = y, n \ge n_0$.

Let $m = \max\{x_i|1 \le i \le 101\}$. Then $x_1^n + 2x_2^n + \dots + 101x_{101}^n$ can be rewritten as follows
$$x_1^n + 2x_2^n + \cdots + 101x_{101}^n = a_m m^n + a_{m-1} (m-1)^n + \cdots + a_1$$
where $a_i \ge 0, \forall i$ and $a_1 + \dots + a_m = 1 + 2 + \dots + 101$. Hence there exists sufficiently large $n$ such that
$$a_m m^n + \dots + a_1 - 1 - y \cdot y^n = 0 \quad (1)$$

The following lemma can help us find $a_i$ and $y$.

Lemma: Given integers $b_1, \cdots, b_N$ and suppose there exists sufficiently large $n$ such that $b_1+b_2 2^n+\cdots b_N N^N = 0$. Then $b_i = 0, \forall i$.

Proof: Suppose some $b_i$ is not 0. Without loss of generality, let $b_N \neq 0$. Dividing $b_1 + b_2 2^n + \cdots b_N N^N$ by $N^N$, we obtain
$$|b_N| = |b_{N-1}|\left(\frac{N-1}{N}\right)^n + \cdots + b_1\left(\frac{1}{N}\right)^n \le (|b_{N-1}| + \cdots + |b_1|)\left(\frac{N-1}{N}\right)^n.$$
For sufficiently large $n$, $\left(\frac{N-1}{N}\right)^n$ is a sufficiently small number, which contradicts $b_N \neq 0$.

Clearly, $y > 1$. Applying the lemma to equation (1), we get $a_m = y = m, a_1 = 1$, and the other $a_i = 0$. From this we obtain $(x_1, \cdots, x_{101}) = (1, m, \cdots, m)$. But we have $1+m = a_1+\cdots+a_m = 1+\cdots+101 = 1+k$ so $m=k$.