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 $n-1$).