Let $f\in \mathbb C[x_1,\dots,x_d]$ be a (nonconstant) polynomial. Of course it can be viewed as a (surjective) regular map $$\tilde f:\mathbb A^d_\mathbb C\to \mathbb A^1_\mathbb C.$$
Question. When is $\tilde f$ proper?
The map $\tilde f$ being finite type and separated, the question asks: when is $\tilde f$ universally closed?
When is it such? Can one say anything through the valuative criterion?
A proper map of affine varieties is finite (Hartshorne, Ex. II.4.6). So the answer is iff $d = 1$.
Here's a proof of ZCN's claim.
Suppose that $X$ and $Y$ are affine varieties, and $f:X\to Y$ is proper. We claim then, that in fact, $f$ is finite. Note that since both $X$ and $Y$ are so well-behaved, we can apply ZMT to see that $f$ is finite if and only if it is quasifinite.
To see that $f$ is quasifinite, we merely note that for each $y\in Y$, the fiber map $X_y\to \text{Spec}(k(y))$ is proper, and since $f$ is affine, that $X_y$ is also affine, say $X_y=\text{Spec}(R)$. By passing to (one of finitely many) components, we may as well assume that $\text{Spec}(R)$ is irreducible. But, note that any $r\in R$ gives us a map $\text{Spec}(R)\to\mathbb{A}^1_{k(y)}$. But, since the image of this map is closed, irreducible, and proper of $\text{Spec}(k(y))$ (with the reduced scheme structure), it must be a point $(p(T))\in k(y)][T]$. This implies that $p(r)=0$, and so $R$ is integral over $k(y)$. Since it's also finite type, this implies that $R$ is finite over $k(y)$, and, in particular, $X_y$ is finite.
You can probably apply the ideas in the second paragraph directly, without using ZMT, but I find this method the most comfortable for me. :)
