Doesn't the definition in the question only require that every nonzero submodule contains a square-root, rather than is a square-root?

I was $(100-\varepsilon)\%$ sure that you intended a supremum in the definition of $\operatorname{ofindim}A$, so I edited to add it. Apologies if I was wrong.

Possibly you mean "generated as a triangulated category with coproducts"? A few years ago, I considered this property, not specifically for commutative rings. I didn't find anything like equivalent conditions on a ring, and it's still quite mysterious when "injectives generate" in this sense.

It's still not clear exectly what you mean by "generated by". You edited to replace "injective modules" by "direct sums of injective modules", but I doubt that's what you mean, as the smallest triangulated subcategory containing all direct sums of injective modules is never $D(R)$, as it is contained in the bounded derived category.

In a general additive category, having a splitting map is not enough to ensure being the inclusion of a direct summand. E.g., in the additive category of real vector spaces with dimension not equal to one, an injective map $\mathbb{R}^2\to\mathbb{R}^3$ has a splitting map, but $\mathbb{R}^2$ is not a direct summand of $\mathbb{R}^3$

In a triangulated category, if $A\to B$ has a splitting map (i.e., a map $B\to A$ such that the composition $A\to B\to A$ is $\operatorname{id}_A$), then it follows from @Dave's comment that $A$ is a direct summand of $B$, as then $C\to\tau A$ is equal to the composition $C\to\tau A\to\tau B\to\tau A$, which is zero, since $C\to\tau A\to\tau B$ is zero, since it is the composition of two consecutive maps in a triangle.

I interpreted the question as "is there a pair $(D,K)$ ...". But more is known for some specific fields, according to Smoktunowicz. For example, there is no such $D$ for $K$ finite or algebraically closed.

@YCor All of which doesn't quite answer your question!

@YCor But if everything she lists as an open question is still open, then I think that "Must every division algebra over a field $K$ that is finitely generated as a division algebra, and such that every element is algebraic over $K$, be finite-dimensional over $K$?" must also be open, as a positive answer would give a positive answer to Kurosh's problem, and a negative answer would give a negative answer to Question 5.

@YCor Smoktunowicz also lists as an open question (Question 5) whether a division ring over a field $K$ that is finitely generated as a $K$-algebra must be finite-fimednsional. But without the condition that every element is algebraic, "finitely generated as an algebra" is stronger than "finitely generated as a division algebra".