I am reading this article
and would like to ask if this section is correct:
If it can, then you would have
$f(x,y) = g(x,y) * h(x,y)$,
where g and h are polynomials of degree at least one (that is, not constants). It turns out that there will necessarily be at least one complex solution $(x,y)$ to the simultaneous equations
$g(x,y) = 0$ $h(x,y) = 0$
This is known from Bezout's Theorem.
Is this true? It seems like $g(x,y) = x^2 + y^2$ and $h(x,y) = x^2 + y^2 + 1$ is an obvious counterexample.
Your example is correct and the proof from your link is not.
The polynomial $2xy + 3x^2 + y$ given there is irreducible just because it is irreducible as a polynomial in $y$ with coefficients in the ring $\mathbb C[x]$.
Sometimes one can use algebraic geometry tools to determine whether a homogeneous polynomial $f$ (in at least three variables) is irreducible: if its zero set is smooth ($f$ and its partial derivatives have no common zeros other than $(0,0,...,0)$), then it is irreducible.