Consider the $m \times 2$ matrix $$M=\begin{pmatrix} \langle \mathbf{x}, \mathbf{z}_1 \rangle^2 & \langle \mathbf{y}, \mathbf{z}_1 \rangle^2 \\ \vdots & \vdots \\ \langle \mathbf{x}, \mathbf{z}_m \rangle^2 & \langle \mathbf{y}, \mathbf{z}_m \rangle^2 \end{pmatrix} $$with $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$ and $m \ge n$.
Is there a way to find out what is the smallest $m$ such that $\text{rank}(M)=2$ for all $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$ linearly independent? Anything can be assumed on the vectors $\mathbf{z}_i$.
I found a specific example where $m=n + \binom{n}{2}$, with the vectors $$\mathbf{z}_1=\mathbf{e}_1, \, \dots \, , \, \mathbf{z}_n = \mathbf{e}_n, \, \mathbf{z}_{n+1}=\mathbf{e}_1+\mathbf{e}_2, \, \dots \, , \, \mathbf{z}_{m} = \mathbf{e}_{n-1} + \mathbf{e}_n,$$with $\mathbf{e}_i$ vectors of the canonical basis of $\mathbb{R}^n$. If a single $\mathbf{z}_i$ is removed, we should have $\text{rank}(M)=1$ (for some selection of $\mathbf{x}$ and $\mathbf{y}$), but I was not able to say much more in the general case...
Thank you!