I am starting by reading a book titled: "Lecture on Module and Ring" by T.Y. Lam, but I have stopped because I don't understand how to obtain that isomorphism that is in the title.
To put a little more into context about what I know, I was reading a little about introduction to modulus theory until I came across something that surprised me and that is that there exist rings $R$ where $R^m\cong R^n$ (as modules) for $n,m$ different, that made me stop my introductory reading and move on to the book I mentioned, so there may be things I don't understand yet.
This is what is mentioned in the book:
Let $V$ be a free right module of infinite range on a ring $k\not = (0)$, and let $R \cong \mathrm{End}_k(V)$. Then, as right $R$ modules, $R^n \cong R^m$ for any natural numbers $n, m$. To do this, it is enough to show that $R \cong R^2$. Fix a $k$-isomorphism $\epsilon \colon V \rightarrow V \oplus V$ and apply the functor $\mathrm{Hom}_k(V, -)$ to this isomorphism. We get an abelian group isomorphism
$$\lambda \colon R\rightarrow \mathrm{Hom}_k (V,V \oplus V)= R \oplus R.$$
We finish by showing that $\lambda$ is a right $R$-module homomorphism. To see this, note that:
$$\lambda(f) = (\pi_1 \circ \epsilon \circ f, \pi_2 \circ \epsilon \circ f) \quad (\forall f \in R),$$
where $\pi_1, \pi_2$ are the two projections of $V \oplus V$ onto $V$.
$V$ is a free module of infinite rank, so it has infinite basis $B$. Because the set $B$ is infinite, we can partition it into two parts, both of which are equinumerous to $B$, so $B = B_1 \cup B_2$, $B_1 \cap B_2 = \varnothing$, and there are bijections $\beta_1 \colon B_1 \to B$, $\beta_2 \colon B_2 \to B$. An element $x \in V$ is a sum $\sum_{b_i \in B} a_i b_i$ for some $a_i \in k$, only finite number of which is nonzero. We define the corresponding $\epsilon(x)$ as the pair $\left(\sum_{b_i \in B_1} a_i \beta_1(b_i), \sum_{b_i \in B_2} a_i \beta_2(b_i)\right)$
To get a specific example, if $V$ is a free module countable rank, then $V$ is a direct sum of countably many copies of $k$, or, equivalently, the set of sequences $(a_n)_{n \in \mathbb{N}}$ with finite support (there is only finite number of indices $i$ such that $a_i \neq 0$). Then an isomorphism $\epsilon\colon V \to V \oplus V$ is given by $\epsilon(a_0, a_1, a_2,\dots) = ((a_0, a_2, \dots), (a_1, a_3, \dots))$ (this is the same construction as before: we split the basis set into two parts with even and odd indices).