Let $K$ be a complex quadratic number field such that $K(i)$ has class number $1$. If $K$ has class number $\ne 1$, then $K(i)$ must be the Hilbert class field of $K$, which, in this case, coincides with the genus class field of $K$. By genus theory, the discriminant of $K$ must have the form $d = -4p$ for a prime number $p \equiv 1 \bmod 4$.
In these cases, the ring of integers in $K$ is $R = {\mathbb Z}[\sqrt{-p}]$, and the ring of integers in $L$ is ${\mathbb Z}[i, (1 + \sqrt{p})/2] \ne R[i]$.
Finding number fields of higher degree with this property seems to be an interesting problem;
I can't think of an obvious approach in general, but if I find anything, I'll let you know.
Edit 1. The argument works for all imaginary number fields: if $R$ is the ring of integers in a number field $K$, and if $S = R[i]$ is the ring of integers in the extension $L = K(i)$, then disc$(L) = \pm 4$ disc$(K)^2$ (this is a simple determinant calculation: take an integral basis
$\{\alpha_1, \ldots, \alpha_n\}$ for $K$; then $\{\alpha_1, \ldots, \alpha_n, i\alpha_1, \ldots, i\alpha_n\}$ is an integral basis for $L$).
On the other hand, if $L$ has class number $1$ and $K$ is not a PID, then $K$ has class number $2$ and $L$ is the Hilbert class field of $K$. This implies disc$(L) = \pm$ disc$(K)^2$.
The problem in the non-imaginary case is that $K(i)$ might be unramified at all finite primes, but not at infinity; in this case, $K$ has class number $2$ in the strict sense, yet its ring of integers is a UFD.
Remark 2. By looking at $p = \alpha^2 + \beta^2$ modulo $4$ it follows (unless I did something stupid) that primes $p \equiv 3, 7 \bmod 20$ are not sums of two squares in
${\mathbb Z}[\sqrt{-5}]$.
Edit 2. Your suggestion to look at rings $R[\frac12]$, where $R$ is the ring of integers of a quadratic number field, seems to work for $K = {\mathbb Q}(\sqrt{-17})$, which has a cyclic class group of order $4$. The ring $R[\frac12]$ has class number $2$ because $2$ is ramified and so generates a class of order $2$, and $S = R[\frac12,i]$ is the integral closure of $R[\frac12]$ in the extension $L = K(i)$. The ring $R[\frac12]$ has class number $2$, and $S$ is a UFD since $L$ has class number $2$, and its class group is generated by one of the prime ideals above $2$ (the ramified prime above $2$ in $K$ splits in $L$).
Edit 3. The corollary concerning sums of two squares in $R[\frac12]$ shows that primes $p \equiv 3 \bmod 4$ splitting in $K$ are sums of two squares up to units. In fact it follows from genus theory that the prime ideals above such $p$ are not in the principal genus, hence lie in the same class as the prime above $2$ or in its inverse. This implies that either $2p = x^2 + 17y^2$ or $8p = x^2 + 17y^2$, giving a representation of $p$ as a sum of two squares up to a unit (remember $2$ is invertible). Thus everything is working fine.
The "Super-$n$" variety, call it $V_n$, seems to be of general type once $n \geq 4$. It probably has no nontrivial rational curves (where "trivial" means that it lies on a hyperplane $c_1=0$ or $c_j=c_k$ some distinct $j,k$; over ${\bf C}$ one must also exclude $c_j=0$ for $j>1$). For $n$ large enough this should follow from the Bombieri-Lang conjectures for the "Super-4" variety $V_4$.
In general if a smooth variety is the "complete intersection" in some projective space ${\bf P}^{N-1}$ of hypersurfaces $P_i=0$ of degrees $d_1,\ldots,d_r$ then it is of general type iff $\sum_i d_i > N$. Here we have $N=(n^2+n)/2$ and $r=(n^2-n)/2$, with each $d_i=2$, so we would get general type once $n\geq 4$; our variety is not quite smooth but the singularities look mild enough not to change the result.
A (very plausible but extremely hard) conjecture of Bombieri and Lang asserts that all the rational points on a variety $X$ of general type lie on a finite union $X_0$ of subvarieties of lower dimension. (This would vastly generalize Faltings' theorems on curves of genus $>1$ [Mordell's conjecture] and subvarieties of abelian varieties.) For complete intersections in ${\bf P}^{N-1}$, the following naïve but suggestive heuristic points in the same way: try the $\sim H^n$ points $(x_1:x_2:...:x_N)$ with integers $x_m$ such that $H \leq \max_m |x_m| < 2H$; for any choice of such $x_m$ we have $|P_i(\vec x)| \ll H^{d_i}$ for each $i$, and if we imagine these $r$ numbers $P_i(\vec x)$ are more-or-less randomly and independently distributed among integers of those sizes then the expected number of $\vec x$ where they're all zero is of order $H^{N-\sum_i d_i}$. This means that the general-type case is precisely when the exponent is negative, and thus that (summing over $H=1,2,4,8,16,\ldots$) the total number of rational points if finite. This heuristic cannot account for non-random rational points due to polynomial identities, but those are precisely the subvarieties that the Bombieri-Lang conjecture allows.
It seems reasonable to guess that already for $n=4$ there are no nontrivial rational curves, and that the nontrivial part of $X_0$ is finite or even empty. Unfortunately, even assuming the B-L conjecture there is no known way to determine $X_0$. Nevertheless it may be possible to deduce that some $V_n$ has no nontrivial points under the assumption that B-L holds for $V_4$. The reason is that for $n > 4$ there are many maps from $V_n$ to $V_4$, obtained by choosing any $4$ of the $n$ variables $c_1,\ldots,c_n$ in order, and the corresponding six $d$'s. If all nontrivial points of $V_4$ were known to lie on a union $X_0$ of proper subvarieties, then any nontrivial point of $V_n$ would have to lie on the intersection of preimages of $X_0$ under $n \choose 4$ different maps, and whatever $X_0$ turns out to be, such an intersection ought to be trivial if $n$ is large enough.
NB it would likely require some nontrivial [sic] work to make a proof of this even assuming the B-L conjecture for $V_4$, but such an analysis was carried out in a similar context in the famous paper
L.Caporaso, J.Harris, and B.Mazur: Uniformity of rational points, J. Amer. Math. Soc. 10 #1 (1997), 1-45
and something like that should be possible here.
Best Answer
For many examples of this kind, see Olga Taussky, "Sums of squares", Amer. Math. Monthly 77 (1970) 805-830.