Rings that satisfy the condition
$R^n \cong R^m \iff n=m$ are said to have *invariant basis number* or the *invariant basis property*. P. M. Cohn has constructed examples of rings which fail to have this property, even giving examples of (non commutative) integral domains for which e.g. $R^3\cong R$ but $R^2\neq R$.

Suppose $R^n\cong R^m$ for some $n,m$. Then there must exist integers $h,k$ such that

$$
R^m\cong R^{m'} \iff m=m'\ \mbox{or}\ m,m'\ge h \ \mbox{and} \ m\equiv m'\ (\mod k)
$$

A ring satisfying this condition is of type $(h,k)$. In [P. M. Cohn, Some remarks on the invariant basis property, Topology 5 (1966), 215-228] a fairly simple construction is given for rings of type $(h,k)$ for any $h,k$.

There are earlier constructions of rings of type $(h,k)$ by Leavitt [W. G. Leavitt:
Modules without invariant basis number, Proc. Am. Math. Sot. 8 (1957), 322-328], [W. G. Leavitt:
The module type of a ring, Trans. Am. Math. Sot. 103 (1962), 113-130], but they are far more complicated.

Certainly we can do algebraic K-theory over rings without the invariant basis property; we just need to be a little more careful. For example we won't necessarily have $K_0(R)\cong {\bf Z} \oplus \tilde{K}_0(R)$.