Bass' stable range for Bezout rings

Question

As discussed in this MO topic, every principal ideal domain has stable rank at most 2. The proof in the accepted answer uses the fact that PID is a unique factorization domain, but there can be no irreducibles in case of Bezout ring (every finitely generated ideal is principal), as shown by the example of the ring of all algebraic integers.

However, if one assumes the ring to be a Bezout domain, one can show that it is Hermitian, and thus of stable rank 2 (this paper, for example).

It looks like there is nothing about infinitely generated ideals in either PID or stable range conditions, so it is naturally to ask if there are Bezout rings of stable rank greater than 2. Surely they have to be non-UFD. The two basic examples I know of: algebraic integers and entire functions, but both are domains (and of stable rank 1, which makes the situation even more puzzling). Another example from the Wikipedia page is of no help either.

Question. Is the stable rank of Bezout rings bounded from above? Can it be strictly greater than 2?

PS. All rings are considered to be commutative.

Answer 1

In "Rings of continuous functions in which every finitely generated ideal is principal" by L. Gillman and M. Henriksen (Trans. Amer. Math. Soc. 82 (1956), 366-391 link), Example 3.4 is of a topological space $X$ such that the ring of continuous functions $C(X)$ is Bezout but not Hermitian. The space $X$ is the complement of a closed half-plane in its Stone–Čech compactification.

Theorem 1 of "Reduction of Matrices over Bezout Rings of Stable Rank not Higher than 2" by B. V. Zabavs'kyi (Ukrainian Mathematical Journal, 55(4) (2003) 665-670 link) states that a commutative Bezout ring is Hermitian if and only if it has stable rank (at most) two.

So putting these together gives a Bezout ring with stable rank greater than two. Because of the remark I made in comments, I guess the stable rank is infinite, but I'm not sure.