Transcendental numbers are a well-known phenomenon: a number $x$ is transcendental if no polynomial with integer coefficients has $x$ as its root, or $p(x)\neq0$ for all polynomials $p$ with integer coefficients.

I was wondering if this concept could be extended to multivariate polynomials:

Is there a point $\vec{x}=(x_1, x_2, \dots, x_n)$ for any $n$ such that $p(\vec{x})\neq0$ for all multinomials $p$ with integer coefficients?

$\pi$ is transcendental, but $(\pi, \pi)$ is a root of $p(x,y)=x-y$, and $(\pi,0)$ is a root of $p(x,y)=x \cdot y$. However, I can't easily find a bivariate polynomial having $(\pi, e)$ as root.