Let $p,q$ be polynomials in $\mathbb{C}[x,y]$ such that the ideal $(p,q)$ is a proper ideal of $\mathbb{C}[x,y]$. Does there exist complex numbers $z,w$ such that $$p(z,w)=0,\,\,\,\,\,q(z,w)=0\ ?$$
Motivation:
Let $K$ be a field and $f$ be a non-constant polynomial in $K[x]$ that does not have a root in $K$. One can choose a non-unit irreducible factor $p$ of $f$ and construct the field extension $K[x]/(p)$. The field resulting will contain an isomorphic copy of $K$ and have a root to $f$.
Now if we try to do this for two variables. Again let $K$ be a field. Let $p,q$ be two polynomials in $K[x,y]$. Now if it happens that $(p,q)=K[x,y]$ then there is no hope of finding a field extension of $K$ that will contain solution to the similtaneous equations $p=0,q=0$ for obvious reasons (by considering the evaluation homomorphism). If $(p,q)$ is a proper ideal of $K[x,y]$ then one can choose a maximal ideal containing $(p,q)$. If we set $E=K[x,y]/I$, one can see easily that the field $E$ contains an isomorphic copy of $K$ and has solutions to the similtaneous equations $p=0,q=0$
Yes:
One of the forms Hilbert's Nullstellensatz (an essential tool in algebraic geometry) is the statement that if $k$ is algebraically closed and $J$ is a maximal ideal of $k[X_1,\dots,X_n]$ then $J = (X_1-x_1,X_2-x_2,\dots,X_n-x_n)$ for some $x_i$ in $k$.
Using the notation $Z(S)$ to denote the points in $k^n$ where all polynomials in set $S$ vanish simultaneously, any proper ideal $I$ is contained in some maximal ideal $J$ of the above form. Then $Z(J)\subset Z(I)$, and clearly $Z(J)$ consists of exactly one point, so $Z(I)$ contains at least this point, i.e. there exists at least one point in $k^n$ where all of the polynomials in $I$ vanish simultaneously.
As a side note, this form of Nullstellensatz is equivalent to several other forms, including the statement $"I(Z(I))=\sqrt{I}"$ which is a very useful tool in determining what polynomials vanish on a given set of points in $k^n$, and is also useful in computing the radical of a given ideal.