Multivariate coprime polynomials in field extensions

Yes, if you let me use some algebraic geometry (it's not really necessary, I also write everything you need in algebraic therms between parentheses, but it really helps intuition).

First, we may suppose that $f$ and $g$ are irreducible, this clearly implies the general statement. Now, we may also suppose $F/E$ algebraic: if not, using Gauss's lemma, we can see that $f$ and $g$ remain prime and relatively prime in $E(\{y_j\}_{j\in J})[x_i]$ (where $\{y_j\}_{j\in J}$ is a trascendence basis), because $E[\{y_j\}_{j\in J}]$ is an UFD.

Now, call $X=\operatorname{Spec~(E[x_i]}/(f,g))$. $X$ (or $E[x_i]/(f,g)$ if you don't like schemes) has dimension $n-2$, because $f$ and $g$ are relatively prime and so $V(f)$ and $V(g)$ have no common components (or $g$ is not a zero-divisor in $E[x_i]/(f)$).

$F/E$ is an algebraic extension, hence $X_F\to X$ (or $F[x_i]/(f,g)\supseteq E[x_i]/(f,g)$) is integral, so it preserves dimension. But if $h\in F[x_i]$ divides both $f$ and $g$, it describes a subscheme of $X_F$ of dimension $n-1$ (or defines a quotient $F[x_i]/(h)$ of $F[x_i]/(f,g)$ of dimension $n-1$).

Let $k$ be a field. We say that a ring or a $k$-algebra is irreducible if its spectrum is irreducible, or equivalently if it has a unique minimal prime. A $k$-algebra is geometrically irreducible if $A\otimes_k k'$ is irreducible for every field extension $k'/k$.

If $k[x_i]$ is a multivariate polynomial ring and $h\in k[x_i]$, then it is easy to see that

(a) $k[x_i]/(h)$ is irreducible iff $h$ is a power of an irreducible polynomial times a unit.

(b) $k[x_i]/(h)$ is reduced iff $h$ is square-free, i.e. $h$ is not divisible by the square of an irreducible polynomial, i.e. $h$ is a product of pairwise non-associate irreducible polynomials.

Now let $F/E$ be a field extension. We present three facts from the Stacks Project.

(1) There exists an intermediate field $E'$ that is a separable algebraic extension of $E$ such that $F$ is a geometrically irreducible $E'$-algebra.

(2) If $F$ is a geometrically irreducible $E'$-algebra and $A$ is an irreducible $E'$-algebra, then $A\otimes_{E'} F$ is irreducible.

(3) If $E'/E$ is a separable field extension and $A$ is a reduced $E$-algebra, then $A\otimes_E E'$ is reduced.

We now begin to prove that if $f,g\in E[x_i]$ are coprime over $E$ then they are also coprime over $F$.

By unique factorization, we may assume that $f$ and $g$ are irreducible over $E$, so $E[x_i]/(fg)$ is a reduced $E$-algebra. Let $E'$ be as in (1), then $E'[x_i]/(fg)$ is reduced by (3), so $f$ and $g$ are coprime over $E'$. To show that $f$ and $g$ are coprime over $F$, we may again assume that $f$ and $g$ are irreducible over $E'$. Then $E'[x_i]/(f)$ and $E'[x_i]/(g)$ are irreducible, so $F[x_i]/(f)$ and $F[x_i]/(g)$ are irreducible by (2), and hence $f=uf'^n$ and $g=vg'^m$ for some $u,v\in F^\times$ and irreducible $f',g'\in F[x_i]$; assume that $n\le m$ without loss of generality. If $f,g$ are not coprime over $F$, then $f'$ and $g'$ are associates and hence $f\mid g$, contradicting that $f,g$ are coprime over $E'$.