Ore and Invariant basis number. A ring $\rm R$ (unitary, associative, non-commutative) has IBN (invariant basis number) if ${\rm R}^n\simeq {\rm R}^m$ (as left-modules) implies $n=m$.
$\rm R$ is left-Ore if $\rm Ra\cap Rb$ is non-zero for all non-zero $(a,b)\in{\rm R}^2$.
Every left-Noetherian ring $R$ both is left-Ore and has IBN.

Question. Is there a left-Ore ring that does not have IBN?

 A: The definition you have given for "left Ore ring" is not the standard one. What you have written is in fact equivalent to $R$ being a left uniform ring, by which I mean that the intersection of any two nonzero left ideals is nonzero.
It's easy to see why such rings always have IBN. Such a ring has left uniform dimension $1$, and $R^n$ would have uniform dimension $n$. But uniform dimension is well-defined, so $R^n\cong R^m$ would imply this module has two different uniform dimensions, a contradiction.

The normal definition of a left Ore ring is this: a ring is called left-Ore if the set of regular elements is left permutable.
The pool of examples of non IBN rings is not well-known, so we don't have much to lose if we check the few known examples.
You probably know that $End_k(V)$ does not have the IBN if $V$ is infinite dimensional over $k$. It's also von Neumann regular since $V$ is a semisimple $k$ module. But all von Neumann regular rings are left and right Ore since they are equal to their rings of classical quotients (see for example Lam's Lectures on modules and rings p 303 in the middle.
