Let $\mathbb{C}[X_1,...,X_d]$ denote the set of polynomials over $\mathbb{C}$ in $d$ variables. Let $\mathcal{A}$ denote the set of all $z \in \mathbb{C}^d$ where there exists $1 \leq i \leq d$ such that $z_i=0$. Let $p \in \mathbb{C}[X_1,...,X_d]$ be such that the zero-set of $p$ is a subset of $\mathcal{A}$, in other words, if $p(X_1,...,X_d)=0$, then $(X_1,...,X_d) \in \mathcal{A}$.
Claim: $p(X_1,...,X_d) = C X_1^{n_1}...X_d^{n_d}$ for some $C \in \mathbb{C}$ and $n_i \in \mathbb{N}_0$
I can not think of a counter example, so I have been trying to prove it using Hilbert's Nullstellensatz, but have not succeeded so far. The problem I face is that I can not come up with an ideal that would both give me my claim as a result, but also has the correct common zero-set. On the other hand, Hilbert's Nullstellensatz seems like quite a heavy theorem to use for a claim that seems more straightforward to me...
The result should be elementarily proven by contraposition. Assume $p$ is not a monomial. Then there exists an index $i_0$ such that $p$ contains monomials of differing degree in $X_{i_0}$. This means that we may write $$p = X_{i_0}^k(X^m_{i_0}p_m+X^{m-1}_{i_0}p_{m-1}+\dots+p_0),$$ where $m\geq 1$, each $p_i$ is a polynomial in $d-1$ variables, and $p_m$ and $p_0$ are non-zero.
Now, if we could find a point $a=(a_1,a_2,\dots,a_{i_0-1},a_{i_0+1},\dots,a_d)\in \mathbb{C}^{d-1}$ such that $p_m(a)\neq 0$, $p_0(a)\neq 0$ and $a\notin\mathcal{A}_{d-1}$, then $P(x) = p(a_1,a_2,\dots,a_{i_0-1},x,a_{i_0+1},\dots,a_d)$ would be a univariate polynomial with a non-zero root, and naming that root $a_{i_0}$, then $\hat{a}=(a_1,a_2,\dots,a_{i_0-1},a_{i_0},a_{i_0+1},\dots,a_n)$ would satisfy $p(\hat{a})=0$ but $\hat{a}\notin\mathcal{A}$, completing the proof by contraposition.
The existence of such a point is simple: We need that $V(p_m)\cup V(p_0)\cup \mathcal{A}_{d-1}$ has a non-empty complement, i.e. this union is not $\mathbb{C}^{d-1}$. But they all have measure zero in $\mathbb{C}^{d-1}$, so the same holds for their union, and so they cannot be $\mathbb{C}^{d-1}$.