Is there a way to prove that $\mathbb P^n \to \operatorname{Spec} \mathbb Z$ is universally closed without the valuative criterion? The title is self explanatory. I was just wondering, since some books such as Hartshorne introduce the valuative criteria basically to show that projective space is proper.
 A: Yes, elimination theory provides an alternate proof. This is mentioned in Hartshorne theorem I.5.7A:

Theorem (Elimination Theory). Let $f_1,\cdots,f_r$ be homogeneous polynomial in $x_0,\cdots,x_n$ having indeterminate coefficients $a_{ij}$. Then there is a set $g_1,\cdots,g_t$ of polynomials in the $a_{ij}$, with integer coefficients, which are homogeneous in the coefficients of each $f_i$ separately, with the following property: for any field $k$, and for any set of special values of the $a_{ij}\in k$, a necessary and sufficient condition for the $f_i$ to have a common zero different from $(0,0,\cdots,0)$ is that the $a_{ij}$ are a common zero of the polynomials $g_j$.
Proof. Van der Waerden, Modern Algebra, vol II, Section 80, p.8.

To show that $\Bbb P^n\to\operatorname{Spec} \Bbb Z$ is universally closed using this theorem, apply the argument found here in the answer to exercise I.6.5. You will need some slight adjustments for the scheme-theoretic case versus the case of varieties, but I think you ought to be capable of filling in the blanks. Please leave a comment if you're looking for a little more.

The full proof that $\Bbb P^n_k\to\operatorname{Spec} k$ is complete has also been presented on MSE before here. The gist of it is that for $Y$ an affine variety and $Z\subset Y\times\Bbb P^n$ a closed subset, we can find generators $\{f_i\}$ of $I(Z)$ which are homogeneous elements of $k[Y][X_0,\cdots,X_n]$ and then express whether $\sqrt{I(Z)}$ contains the irrelevant ideal by a determinantal condition derived from the multiplication map $\bigoplus P_{d-\deg f_i} \to k[Y][X_0,\cdots,X_n]_d$, $(p_i)\mapsto \sum f_ip_i$ which only involves elements of $k[Y]$.
