For which fields $k$ does the following hold for all $n \geq 1$? Let $(a_1,\ldots,a_n) \in k^n$. Then there exists a polynomial $f(x_1,\ldots,x_n) \in k[x_1,\ldots,x_n]$ such that $f(a_1,\ldots,a_n) = 0$ but $f(b_1,\ldots,b_n) \neq 0$ for all $(b_1,\ldots,b_n) \neq (a_1,\ldots,a_n)$?
This is clearly impossible for $k$ algebraically closed.
It does hold for $k$ a subfield of $\mathbb{R}$; indeed, in that case we can take
$$f(x_1,\ldots,x_n) = (x_1-a_1)^2 + (x_2-a_2)^2 + \cdots + (x_n-a_n)^2.$$
If $k$ is not algebraically closed, such a polynomial always exists (the opposite is also true and is mentioned in the post).
We may assume that $a_i=0$ for all $i$. Take an irreducible polynomial $g(x)$ of degree $d>1$, then for the homogeneous form $G(x,y)=y^dg(x/y)$ we have $G(x,y)=0$ if only if $x=y=0$. This solves the case $n=2$, for $n=3$ consider the polynomial $G(G(x,y),z)$, it takes zero value only when $x=y=z=0$, and so on.
I guess finite fields $k = \mathbb{F}_{q}$ satisfy this property, namely we can take $$ f = ((x_{1}-a_{1})^{q-1}-1) \dotsb ((x_{n}-a_{n})^{q-1}-1) - (-1)^{n} $$ for any $n$.
In fact we have a Lagrange interpolation formula for finite fields, see this answer.
