In the extension $\mathbb{Q}(\sqrt{-5}, i)/\mathbb{Q}(\sqrt{-5})$, must principal prime ideals of $\mathbb{Z}[\sqrt{-5}]$ necessarily split into 2? Must nonprincipal prime ideals not split?
Rephrased, does a nonzero prime ideal of $\mathbb{Z}[\sqrt{-5}]$ split completely in $\mathbb{Q}(\sqrt{-5},i)/\mathbb{Q}(\sqrt{-5})$ if and only if it is a principal ideal?
I am looking for a more elementary argument that hopefully doesn't invoke class field theory...
Yes, this is true. It is a consequence of class field theory. Indeed, $H=\mathbb{Q}(\sqrt{-5},i)$ is the Hilbert class field of $K=\mathbb{Q}(\sqrt{-5})$. Thus, by class field theory, a prime of $K$ is principal if and only if it splits completely in $H/K$.
More generally, the Hilbert class field $H$ of a number field $K$ enjoys the following property:
In the case of $K=\mathbb{Q}(\sqrt{-5})$, we have $h_k=2$, and $H/K$ is quadratic. If $\wp$ is principal, then $f=1$, and therefore there are $2/1=2$ primes of $H$ above $\wp$. Since $H/K$ is quadratic, that means $\wp$ is completely split. Conversely, if $\wp$ is non-principal, then $f=2$, and so there are $2/2=1$ primes of $H$ above $\wp$, i.e., $\wp$ is inert.
You can give a somewhat ad hoc proof because $\mathbb{Q}(\sqrt{-5},i)\subset \mathbb{Q}(\zeta_{20})$. The splitting behaviour depends on congruences mod $20$, and one can then relate principality to congruences as well by norm equations, and then argue that (miraculously) the principal primes and split primes give exactly the same congruences.
Write $K=\mathbb{Q}(\sqrt{-5})$ and $L=K(i)$. Then $\mathcal{O}_K=\mathbb{Z}[\sqrt{-5}]$ and we have unique primes ramified prime $\mathfrak{p}_2=(2,1+\sqrt{-5})$ and $\mathfrak{p}_5=(\sqrt{-5})$ over $2$ and $5$ respectively. I leave it to you to verify that the extension $L/K$ is unramified, and that for any prime $\mathfrak{p}$ of $K$ lying over the rational prime $p$ of $\mathbb{Q}$ we have the following splitting behaviour:
One can verify these statements by computing all the decomposition fields for primes $p$ in $\mathbb{Q}(\zeta_{20})$. A useful remark for this: a rational prime $p$ cannot be inert in $L$, for then one would have a cyclic decomposition group of order $4$ inside the non-cyclic Galois group of $L/\mathbb{Q}$.
With this i know claim that $\mathfrak{p}$ splits in $L$ if and only if it is principal.
When $p=2$ we have $\mathfrak{p}=\mathfrak{p}_2$, which is non-principal. It it also inert in $L$ so the claim holds for $p=2$. Similarly the claim holds for $p=5$ as $\mathfrak{p}_5$ is principal and it splits in $L$.
Now if $p$ is inert in $K$, we have that $\mathfrak{p}=(p)$ is clearly principal, and it also splits, so the claim holds in this case. Thus for the remainder of the argument we can assume that $p$ is split in $K$, so that $\mathfrak{p}$ has norm $p$.
Assume that $\mathfrak{p}$ is principal, say $\mathfrak{p}=(a)$. Then there exist $x,y\in\mathbb{Z}$ such that
$$ p=N(\mathfrak{p})=N_{K/\mathbb{Q}}(a)=x^2+5y^2. $$
Reducing this mod $4$ we see that $p\equiv 1\bmod 4$, so comparing with the splitting behaviour we see that $\mathfrak{p}$ must split in $L$.
Now suppose that $\mathfrak{p}$ is not principal. Then using that $Cl_{K}$ has order $2$, and that $\mathfrak{p}_2$ is also non-principal, we must have that the product $\mathfrak{p}_2\mathfrak{p}$ is principal, so similarly we end up with an equation of the form
$$ 2p=N(\mathfrak{p}_2\mathfrak{p})=x^2+5y^2 $$
for certain $x,y\in\mathbb{Z}$. Now if $p\equiv 1\bmod 4$, this would yield $x^2+5y^2\equiv 2\bmod 8$, which cannot happen. Thus $p\equiv 3\bmod 4$, so that $p\equiv 3,7\bmod 20$, and we see from the splitting behaviour that $\mathfrak{p}$ is inert in $L$.
