My favorite version of ZMT is the same as Francesco's: Let $f \colon X \to Y$ be a birational projective morphism of noetherian integral schemes, with $Y$ normal. Then the fibre $f^{-1}(y)$ is connected for any $y \in Y$.
Let me try to give a pedestrian${}^1$ pedestrian naive${}^1$ answer:
If $f^{-1}(y)$ is not connected, then $f$ "glues" two or more pieces together. In particular, and this is the key, it follows that image point is not unibranched or in other words its tangent cone is not irreducible. Therefore a local equation for $Y$ would look something like $$uv+ p=0.$$ p=0,$$ where $p$ has a higher degree (in whatever sense locally at the point in question) than $uv$. Now for simplicity assume that $p$ is a polynomial of $u,v$. This is not entirely true, but I am not claiming to prove ZMT here. At least it is true for a nodal cubic.
Anyway, once we have that, we're kind of done: if $uv + p(u,v)=0$, divide by $av^{d+N}$ where $av^du^N$ is the highest degree term of $p$ in $u$ and obtain a monic polynomial of degree $N$ in $\dfrac uv$.
${}^1$: pedestrian naive = heuristic, not trying to be precise
