I believe this follows using Vandermonde matrices. First I will prove a proposition that gives a bit more than what the OP asks.

For every integer $n \geq 2$, for every integer $r=1,\dots,n$, define
$$ F_r(X_1,\dots,X_n) = F_{n,r}(m_1,\dots,m_n;X_1,\dots,X_n) = m_1X_1^r + \dots + m_nX_n^r.$$ Let $P\subset \mathbb{C}$ be a nonempty subset that is stable under addition and which does not contain $0$, e.g., the set of positive integers.

**Proposition $Q_n$.** For every integer $n\geq2$, for every choice of $m_1,\dots,m_n$ in $P$, every nonzero solution $(A_1,\dots,A_n)\neq (0,\dots,0)$ of $$
F_{n,1}(m_1,\dots,m_n;X_1,\dots,X_n)=\dots=F_{n,n-1}(m_1,\dots,m_n;X_1,\dots,X_n)=0$$ is not a solution of $F_{n,n}(m_1,\dots,m_n;X_1,\dots,X_n) = 0$, it has no coordinate equal to $0$, and it has no repeated coordinates.

**Proof** This will be proved by induction on $n$. It is straightforward to prove $Q_2$, the base case of the induction. Basically it boils down to the fact that none of $m_1$, $m_2$ nor $m_1+m_2$ can equal $0$.

Now by way of induction, assume that $n>2$ and that $Q_m$ holds for all integers $2\leq m < n$. Let $(A_1,\dots,A_n)$ be a nonzero solution of the system $F_1=\dots=F_{n-1}=0$. If any $A_i$ is zero, then after permuting the coordinates and $(m_1,\dots,m_n)$, we may assume that $A_n=0$.
Then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system
$$ F_{n-1,1}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,\dots ,$$
$$ F_{n-1,n-2}(m_1,\dots,m_{n-1};X_1,\dots,X_{n-1}) = 0,$$
and which also solves the equation $F_{n-1,n-1}=0$. This contradicts the induction hypothesis. Thus, no $A_i$ equals $0$.

Similarly, if any coordinates are repeated, say $A_{n-1}=A_n$, then $(A_1,\dots,A_{n-1})$ is a nonzero solution of the system,
$$ F_{n-1,1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0, \dots, $$
$$ F_{n-1,n-2}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1}) = 0,$$
and which also solves the equation $$F_{n-1,n-1}(m_1,\dots,m_{n-2},m_{n-1}+m_n;X_1,\dots,X_{n-1})=0.$$ Again this contradicts the induction hypothesis. Thus, there are no repeated coordinates.

Therefore, let $(A_1,\dots,A_n)$ be an $n$-tuple with no zero coordinates and no repeated coordinates. Then, by the theory of the Vandermonde determinant, there is only the trivial solution $(Y_1,\dots,Y_n)$ of the following linear system,
$$
\left\{ \begin{array}{rrrrrrr}
Y_1A_1 & + & \dots & + & Y_nA_n & = & 0, \\
Y_1A_1^2 & + & \dots & + & Y_nA_n^2 & = & 0, \\
\vdots & & \ddots & & \vdots & & \vdots \\
Y_1A_i^r & + & \dots & + & Y_nA_n^r & = & 0, \\
\vdots & & \ddots & & \vdots & & \vdots \\
Y_1A_1^n & + & \dots & + & Y_nA_n^n & = & 0.
\end{array} \right.$$ Indeed, the determinant of the coefficient matrix is
$$
\text{det}([A_s^r]_{1\leq r,s \leq n}) = \pm A_1\cdots A_n \prod_{k<l}(A_l-A_k),
$$
which is nonzero since all $A_i$ are nonzero and distinct. Therefore, if $(m_1,\dots,m_n)$ is a solution, then all $m_i$ equal zero. But this contradicts the hypothesis that all $m_i$ lie in $P$, which does not contain $0$. Therefore, for $(A_1,\dots,A_n)$ as above, there is no choice of $(m_1,\dots,m_n)$ in $P$ such that all $F_{n,1} = \dots = F_{n,n-1}=F_{n,n}=0$. Therefore, if $(A_1,\dots,A_n)$ is as above and solves $F_{n,1}=\dots=F_{n,n-1}=0$, then it does **not** solve $F_{n,n}=0$. Therefore Property $Q_n$ holds. Therefore, for every integer $n\geq 2$, Property $Q_n$ holds by induction on $n$.

**End Proof of Proposition $Q_n$.**

Combinining the proposition with Francesco's homogenization argument, there are $n!$ solutions of the original system, when counted with multiplicity. The point is that, for Francesco's homogeneous system, $x_0$ **cannot** equal $0$, or that would force at least one of the coordinates $A_i$ to be zero, contradicting the proposition.

But, in fact, the multiplicity is always $1$. If you form the Jacobian matrix of Francesco's system evaluated at a solution $(A_0,A_1,\dots,A_n)$, it is the following $n\times(n+1)$-matrix,
$$
\left[
\begin{array}{rr}
-nA_0^{n-1} & * & * & \dots & * \\
0 & m_1 & m_2 & \dots & m_n \\
0 & 2m_1A_1 & 2m_2A_2 & \dots & 2m_n A_n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & rm_1A_1^{r-1} & rm_2A_2^{r-1} & \dots & rm_nA_n^{r-1} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & (n-1)m_1A_1^{n-2} & (n-1)m_2A_2^{n-2} & \dots & (n-1)m_nA_n^{n-2}
\end{array} \right].
$$
If you delete any one of the final $n$ columns, then the determinant of the remaining square matrix is,
$$
\pm n!m_1\cdots m_n A_0^{n-1}\prod_{k<l} (A'_l-A'_k),
$$
where $(A'_1,\dots,A'_{n-1})$ is what you get from $(A_1,\dots,A_n)$ by removing the entry corresponding to the deleted column. By the proposition, this determinant is nonzero. Thus the Jacobian matrix has full rank $n$ at every solution of Francesco's system. Therefore every solution has multiplicity $1$.

In summary, for every choice of $(m_1,\dots,m_n)$ in $P$, there are precisely $n!$ solutions of the original system.

**Edit.** In fact, the argument above holds so long as $(m_1,\dots,m_n)$ is a sequence of complex numbers such that no "subset sum" equals $0$, i.e., $m_{i_1} + \dots + m_{i_q} \neq 0$ for every $1\leq i_1 < \dots < i_q \leq n$. Since Francesco's homogeneous system defines a closed subscheme $Z$ of the product of projective spaces $\mathbb{P}^{n-1} \times \mathbb{P}^n$ with homogeneous coordinates $([m_1,\dots,m_n],[X_0,X_1,\dots,X_n])$, and since the projection $\pi_1:Z\to \mathbb{P}^{n-1}$ is finite, étale of degree $n!$ on the complement of the union of hyperplanes $Z(m_{i_1} + \dots + m_{i_q})$, then presumably the branch locus is precisely this union of hyperplanes.