In what other ways can this chain of ideals ascend? A couple of months ago I asked about non-Noetherian domains, and was given this excellent example of an infinitely ascending chain of ideals:
$$\ldots \subset \langle 16 \rangle \subset   \langle 8 \rangle \subset \langle 4 \rangle \subset \langle 2 \rangle \subset \langle \sqrt{2} \rangle \subset \langle \sqrt[4]{2} \rangle \subset \langle \sqrt[8]{2} \rangle \subset \langle \sqrt[16]{2} \rangle \subset \dots$$
It occurs to me that if we're in the ring of all algebraic integers, we can go to $\langle 2 \rangle$ and there is a different way this chain can ascend:
$$\ldots \subset \langle 16 \rangle \subset  \langle 8 \rangle \subset \langle 4 \rangle \subset \langle 2 \rangle \subset \langle \sqrt{-2} \rangle \subset \langle \sqrt[4]{-2} \rangle \subset \langle \sqrt[8]{-2} \rangle \subset \langle \sqrt[16]{-2} \rangle \subset \dots$$
Since $\langle -2 \rangle = \langle 2 \rangle$, right?
From $\langle 2 \rangle$, are there other ways this chain can ascend? What about from other links? Are these other possibilities "tight" in the sense that if we try to "insert" a link, the new link is equal to either the next lowest or next highest link?
Edit: It was pointed out in the comments that the two chains are in fact identical. As for the second part of my question, regarding "tightness," I may still be laboring under a misunderstanding or it might be a perfectly valid question: is it possible to insert new links into the example chain? What about other infinitely ascending chains? Are tight chains possible in non-Noetherian domains?
 A: What about from other links? Are these other possibilities "tight" in the sense that if we try to "insert" a link, the new link is equal to either the next lowest or next highest link?
You can insert new links anywhere in your chain. For example, between $\langle 2\rangle$ and $\langle {2}^{\frac{1}{2}}\rangle$ you can insert $\langle 2^{\frac{3}{4}}\rangle$. Indeed, for any nonnegative rational number $r=\frac{p}{q}$ the number $\alpha=2^{r}$ is an algebraic integer (it is a root of $x^q-2^p$), and the set $\{\langle 2^{r}\rangle\;|\;r\in\mathbb Q_{\geq 0}\}$ of principal ideals is a chain containing yours, which is ordered like the reverse of the order on the nonnegative rational numbers. ($\langle 2^{r}\rangle\subsetneq \langle 2^{s}\rangle$ iff $s<r$.)
If you do not restrict the members of your chain to principal ideals, you can create a chain of ring ideals ordered like the lattice of order filters of $\langle \mathbb Q_{\geq 0};\leq\rangle$. (An order filter is just an up-closed subset.) Namely, associate to an order filter 
$F\subseteq \mathbb Q_{\geq 0}$ the ideal generated by all numbers $2^r$, $r\in F$. This chain of ideals contains yours and has size continuum.
Are tight chains possible in non-Noetherian domains?
Any chain can be enlarged to a maximal chain, by the Hausdorff maximal principle. Any maximal chain is "tight" in that it cannot be further refined.
However, there must be "gaps" in any such chain, which appear to admit the possibility of inserting a new element, yet do not actually admit insertion. Reason: the ideal lattice of a ring is an algebraic lattice (meaning: it is complete and compactly generated.) If $M$ is a maximal chain in an algebraic lattice $L$, then $M$ is an algebraic chain. By algebraicity, between any two elements $a<b$ of $M$ there must exist a covering pair, which is a pair $c<d$ such that $a\leq c<d\leq b$ with no intermediate element $c<e<d$.  And even though the covering pairs occur densely throughout $M$, so there appear to be places to insert new links, one cannot actually insert new links in a maximal chain. 
On the other hand, if you are wondering whether a non-Noetherian domain can have a maximal chain where each element has an immediate successor and an immediate predecessor, as in your example, the answer is No. Any algebraic chain with this property must be finite, and a non-Noetherian domain has no finite maximal chains in its ideal lattice.
