Let $E$ be simple $R$-module, suppose exists $n,m$ positive integers, $E^{(n)} \stackrel{\phi}{\cong} E^{(m)}$ (as $R$-modules) then $n=m$.
The way Lang does it is by saying
$M_n(End_{R}(E)) \cong End_{R}(E^{(n)})=End_{R}(E^{(m)}) \cong M_m(End_{R}(E))$. Noting that $End_{R}(E)$ is a division ring, and dimension is well defined.
I was thinking of the following: Lable LHS, RHS by $E_1,\ldots, E_n$. wlog, $n>m$.
$\phi_{ij}:=\pi_j \circ \phi \circ j_i:E_i \rightarrow E_j$ is either $0$ or isomoprhism from Schur's lemma. $j_i, \pi_j$ are the inclusion and projection maps respectively.
As $\phi$ is iso, must exists $j$ such that $\phi_{ij}$ is an iso. By pigeonhole princple, exists $1 \le i \not=l \le n$, such that $E_i$, $E_l$ have image on $E_j$ under $\phi$.
$\phi$ is not injective. Contradiction. $n \le m$. By symmetry, $n=m$.
I wonder if this is a valid proof.
Your way is not valid. You can't apply the pigeonhole principle because $E^{(n)}$ generally has more than $n$ simple submodules. (It's just that only $n$ of them appear in any given decomposition, but that's not enough.)
Lang's proof appears to be valid, but to my mind it seems an odd way to prove a special case of the Krull-Schmidt theorem.