Is there a simple way of proving that $\text{GL}_n(R) \not\cong \text{GL}_m(R)$? Letting $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ be fields, and letting $n \geq 3$ and $m$ be natural numbers, it is known that $\text{GL}_{m}(\mathbb{F}_{1})$ and $\text{GL}_{n}(\mathbb{F}_{2})$ are elementarily equivalent if and only if $m=n$ and $\mathbb{F}_{1} \equiv \mathbb{F}_{2}$ (as proven in "Elementary Properties of Linear Groups" in the collection "The Metamathematics of Algebraic Systems — Collected Papers: 1936–1967").
So, given a field $\mathbb{F}$, if $n \neq m$, then $\text{GL}_{m}(\mathbb{F}) \not\equiv \text{GL}_{n}(\mathbb{F})$, and thus $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$.
Letting $R$ be a commutative ring (with unity), and letting $n, m \in \mathbb{N}$ be such that $n \neq m$, is there a simple "algebraic" way of proving that $\text{GL}_{m}(R)$ and $\text{GL}_{n}(R)$ are not isomorphic (as groups)? Is there a simple group-theoretic way of showing that $\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})$ for a field $\mathbb{F}$?
Certain special cases of this problem trivially hold, for example in the case whereby $\mathbb{F}$ is finite, in which case $|\text{GL}_{m}(\mathbb{F})| \neq |\text{GL}_{n}(\mathbb{F})|$.

If the ring is a field  $F$ with characteristic not equal to $2$, then it is not hard to prove that the largest $k$ with $C_2^k \le {\rm GL}(n,F)$ is $k=n$.
We prove this by induction on $n$. Since $-1$ is the only element of order $2$ in $F$, the result holds for $n=1$, so assume that $n>1$. Then $k>1$, so we can choose an element $x$ in this subgroup with $x \ne -I_n$. Then $x$  is conjugate to a diagonal matrix with $1$s and $-1$s on the diagonal, and its centralizer is isomorphic to ${\rm GL}(m,F) \times {\rm GL}(n-m,F)$ for some $m$ with $1 \le m < n$. Now the result follows by induction.
I don't know whether there is an equally elementary argument for fields of characteristic $2$.
