An on-going effort to answer this question has brought up various other questions. Most recently, I have been wondering: given a set of $n$ distinct non-zero complex numbers, $z_1, \dots, z_j, \dots , z_n$, and given the values of the first $n+1$ the power sums, $\mathcal{Z}_0, \mathcal{Z}_1, \dots, \mathcal{Z}_n$: $\begin{bmatrix} 1 & 1 & \cdots & 1 \\ z_{1} & z_{2} & \cdots & z_{n} \\ z_{1}^{2} & z_{2}^2 & \cdots & z_{n}^{2} \\ \vdots & \vdots & \ddots & \vdots \\ z_{1}^n & z_{2}^n & \cdots & z_{n}^n \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = \begin{bmatrix} \mathcal{Z}_0 \\ \mathcal{Z}_1 \\ \mathcal{Z}_2 \\ \vdots \\ \mathcal{Z}_n \end{bmatrix} $
from the Linear Algebra of Symmetric Sums question, we know that these $n$ equations (if you don't include $\mathcal{Z}_0 = n$) determine the $z_j$'s. What if we consider the converse? How does requiring the $z_j$'s to be non-zero and distinct restrict the possible values for the power sums, the $\mathcal{Z}_k$'s? Clearly, they cannot all be zero, but there must be a large number of other choices for $\mathcal{Z}_1, \dots, \mathcal{Z}_n$ that produce non-distinct or zero values for some of the $z_j$'s.
Note
This may appear to ask two different questions:
What is the effect of requiring the $z_j$'s to be non-zero?
What is the effect of requiring them to be distinct?
But a $z_j$ that is zero or two, say $z_j$ and $z_l$, that are the same have a very similar effect on the power sum equations: they shrink the problem from a problem with $n$ independent equations to one with $n-1$ (or fewer) independent equations and an extra dependent equation or two.
In fact, considering this question in this fashion, I may have a sort of answer, albeit a horrendously ugly answer. Consider each of the myriad, but finite number of, ways the $z_j$'s can be degenerate: one $z_j$ zero, two of them zero, one zero and two not distinct, etc., and with that as an assumption, using the technique of the second half of this answer, attempt to calculate the dependent $\mathcal{Z}_k$ equations from the others. If it works, the present set of $\mathcal{Z}_k$'s is degenerate.
Anyway, I am looking for a more tractable answer which might allow me to look at some sets of values for the $\mathcal{Z}_k$'s (especially sets with lots of zeros) and recognize them as not possible.
A Reformulation of the Question
Given the discussion in the note, or given Arturo's determinant calculation in his comment, we can reformulate the question equivalently as:
Given a set of any $n$ complex numbers, $z_1, \dots , z_n$, and given the values of the first $n+1$ the power sums, $\mathcal{Z}_0, \mathcal{Z}_1, \dots, \mathcal{Z}_n$:
$ \mathcal{Z}_0 =n $ $\begin{bmatrix} z_{1} & z_{2} & \cdots & z_{n} \\ z_{1}^{2} & z_{2}^2 & \cdots & z_{n}^{2} \\ \vdots & \vdots & \ddots & \vdots \\ z_{1}^n & z_{2}^n & \cdots & z_{n}^n \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = \begin{bmatrix} \mathcal{Z}_1 \\ \mathcal{Z}_2 \\ \vdots \\ \mathcal{Z}_n \end{bmatrix} ,$ can you use the values of the power sums to determine whether the matrix is singular or not?