Let $F\subseteq \mathbf{Z}^2$, ($\mathbf{Z}^2$ is the integer lattice), such that $|F| = \alpha < \infty$. For any $(m_1,m_2)\in \mathbf{Z}^2\setminus F$, we can define a double trigonometric polynomial. Let $p_{(m_1,m_2)}(x,y)\in L^2([0,1]^2)$ be defined as follows:
$$p_{(m_1,m_2)}(x,y) = e^{2\pi i m_1 x}e^{-2\pi i m_2 y} + \sum_{(n,k)\in F}c_{(m_1, m_2),(n,k)}e^{2\pi i n x}e^{-2\pi i k y}.$$
where the coefficients $c_{(m_1,m_2),(n,k)}$ can change depending on the specific $(m_1,m_2)$ chosen. Further let $Z_{(m_1,m_2)}$ be the zero set of $p_{(m_1,m_2)}$. In other words,
$$Z_{(m_1,m_2)} = \{(x,y)\in [0,1]^2\;:\;p_{(m_1,m_2)}(x,y)=0\}.$$
Here's my following conjecture.
$$ \left|\bigcap_{(m_1,m_2)\not\in F}Z_{(m_1,m_2)}\right| \leq |F| = \alpha.$$
In other words, the sequence of double trigonometric polynomials $\{p_{(m_1,m_2)}\}_{(m_1,m_2)\not\in F}$ have at most $\alpha$ shared zeros. Note that if we can prove the following claim, then our result follows.
Let $\{(x_1,y_1), \ldots, (x_{\alpha+1},y_{\alpha+1})\}\subseteq [0,1]^2$. Then exists some $(m_1, m_2) \in \mathbf{Z}^2\setminus F$ such that the following system of equations is inconsistent: $$ \sum_{(n,k)\in F}c_{(m_1, m_2),(n,k)}e^{2\pi i n x_j}e^{-2\pi i k y_j} = -e^{2\pi i m_1 x_j}e^{-2\pi i m_2 y_j},\quad j = 1, \ldots, \alpha+1. $$
One can see that our system is basically a generalized Vandermonde matrix with complex entries. However, not much is known about generalized Vandermonde matrices with complex entries. My intuition is saying that this must be true as we have more equations than unknowns. Hence, there must be some $(m_1, m_2)$ such that the vector $$(-e^{2\pi i m_1 x_1}e^{-2\pi i m_2 y_1}, \ldots, -e^{2\pi i m_1 x_{\alpha+1}}e^{-2\pi i m_2 y_{\alpha+1}} )^T$$ is not contained in the span of the column vectors of the coefficient matrix.
I can prove that this fact holds for $|F| = 1,2$, but after that the matrices become unwieldy. Any advice or direction would be appreciate or if someone can cook up a counterexample.
For sake of notation, let $Q = [0,1]\times [0,1]$. Additionally, let $$ \delta_{x_i\xi_i} = \delta(x-x_i, \xi-\xi_i), $$ where $\delta$ is the traditional dirac delta functional. Additionally, we have that $$ \langle f, \delta_{x_i\xi_i} \rangle = f(x_i, \xi_i). $$
The desired result follows immediately from the following lemma.
Lemma. Let $(x_1, \xi_1),\ldots, (x_k,\xi_k) \in Q$. Set $$ v_{mn} = \begin{bmatrix} e^{2\pi i m x_1}e^{-2\pi i n \xi_1} \\ \vdots \\ e^{2\pi i m x_k}e^{-2\pi i n \xi_k} \end{bmatrix}, \qquad m,n\in \mathbf{Z}. $$ Then $\{v_{mn}\}_{m,n\in Z}$ is complete in $\mathbf{C}^k$.
Proof. Assume that ${a} = (a_1, \ldots, a_k)\in \mathbf{C}^k$ is orthogonal to $v_{mn}$ for every $m,n\in \mathbf{Z}$. Then for each $m,n\in \mathbf{Z}$ we have that \begin{align*} 0 &= v_{mn}\cdot a \\ &= \sum_{j=1}^{k}\overline{a}_j e^{2\pi i m x_j}e^{-2\pi i n \xi_j} \\ &= \sum_{j=1}^k \overline{a}_j \langle e^{2\pi i m x}e^{-2\pi i n \xi}, \delta_{x_j\xi_j}\rangle \\ &= \left\langle e^{2\pi i m x}e^{-2\pi i n \xi},\sum_{j=1}^k a_j\delta_{x_j\xi_j} \right\rangle. \end{align*}
Note that $\sum_{j=1}^k a_j\delta_{x_j\xi_j} \in C[0,1]'$ and therefore it is uniquely determined by its Fourier Coefficients. Therefore, it follows that $\sum_{j=1}^k a_j\delta_{x_j\xi_j} = 0$ in a distributional sense on $C[0,1]$. However, one can show that $\{\delta_{x_j,\xi_j}\}$ is a linearly independent set. Hence, the coefficients $a_j = 0$ for all $j$. Thus, $\{v_{mn}\}_{m,n\in Z}$ is complete in $\mathbf{C}^k$.
Now, if the sequence of functions $\{p_m\}_{m\not\in F}$ had more than $|F|$ common zeros then that would imply that there exists a set of $|F|-1$ vectors that spans $\mathbf{C}^{|F|}$, which is a contradiction.