A. I. Mal'cev proved the following remarkable result concerning the elementary equivalence of general linear groups: given fields $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ and natural numbers $m$ and $n \geq 3$, $\text{GL}_{m}(\mathbb{F}_{1})$ and $\text{GL}_{n}(\mathbb{F}_{2})$ are elementarily equivalent if and only if $m = n$ and $\mathbb{F}_{1} \equiv \mathbb{F}_{2}$.
Given elementarily inequivalent fields $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$, there are often a variety of simple ways of proving that $\text{GL}_{m}(\mathbb{F}_{1}) \not\equiv \text{GL}_{n}(\mathbb{F}_{2})$. To prove that these groups are not elementarily equivalent, it is natural to consider the centers of these groups. Given the predicate $$\text{Cen}(z) := \forall g \ zg = gz,$$ it is clear that the problem of proving that $\text{GL}_{m}(\mathbb{F}_{1}) \not\equiv \text{GL}_{n}(\mathbb{F}_{2})$ reduces to the 'easier' problem of proving that $\mathbb{F}_{1}^{\ast} \not\equiv \mathbb{F}_{2}^{\ast}$. For example, the sentence $$\exists x \ \text{Cen}(x), x^4 = 1, x^2 \neq 1$$ may be used to prove that $\text{GL}_{m}(\mathbb{C})$ and $\text{GL}_{n}(\mathbb{R})$ are not elementarily equivalent.
I previously asked whether or not it is true that $\mathbb{F}_{1} \not\equiv \mathbb{F}_{2}$ implies $\mathbb{F}_{1}^{\ast} \not\equiv \mathbb{F}_{2}^{\ast}$. The following counterexample was given: $\mathbb{Q} \not \equiv \mathbb{Q}(\sqrt{2})$ since the sentence $\exists x \ x^2 = 1+1$ is true in $\mathbb{Q}(\sqrt{2})$ but not in $\mathbb{Q}$, but $\mathbb{Q}^{\ast} \equiv \mathbb{Q}(\sqrt{2})^{\ast}$ since $\mathbb{Q}^{\ast} \cong \mathbb{Q}(\sqrt{2})^{\ast}$ (by unique factorization of primes).
Similarly, $\mathbb{Q}(\sqrt{3}) \not\equiv \mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})^{\ast} \equiv \mathbb{Q}(\sqrt{2})^{\ast}$. Therefore, the strategy described above involving the centers of general linear groups cannot be used to prove that $$\text{GL}_{m}(\mathbb{Q}(\sqrt{3})) \not \equiv \text{GL}_{n}(\mathbb{Q}(\sqrt{2})) $$ since the centers of these groups are elementarily equivalent.
Letting $L = L_{\text{group}}$ denote the language of groups, it thus seems natural to ask: what is a simple example of an $L$-sentence $\sigma$ such that $\text{GL}_{n}(\mathbb{Q}(\sqrt{3})) \models \sigma$ and $\text{GL}_{n}(\mathbb{Q}(\sqrt{2})) \not\models \sigma$?
Here is an approach to doing this, although I don't want to write down the sentence explicitly.
You can identify the dimension $n$ using just first order language as the largest $k$ such that $C_2^k \le G$, so the first order theories are different for different dimensions, and we can assume that $m=n$. This argument works for all fields with characteristic not equal to $2$.
Let's consider the case $n=2$ first, so we want to distinguish between ${\rm GL}_2(F_1)$ and ${\rm GL}_2(F_2)$ where $F_1 = {\mathbb Q}(\sqrt{2})$ and $F_2 = {\mathbb Q}(\sqrt{3})$ using first order language.
A matrix in either of these two groups is conjugate to one of $$\left(\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right),\ \left(\begin{array}{cc}\lambda&\mu\\0&\lambda\end{array}\right),\ \left(\begin{array}{cc}\lambda_1&0\\0&\lambda_2\end{array}\right),$$ for some $\lambda,\mu \ne 0$, $\lambda_1\ne \lambda_2$.
By considering their eigenvalues, we see that the only non-identity elements that could be conjugate to their squares are those of the second type with $\lambda=1$, which I will call $A_\mu$.
Furthermore any matrix $C$ that conjugated $A_\mu$ to its square $A_\mu^2=A_{2\mu}$ would have to preserve their common eigenspace of $A$ and $A_\mu$, and so would have to be upper triangular, and then you can assume $C$ is diagonal by multiplying it by a power of $A_\mu$. Then, if $C=\left(\begin{array}{cc}\alpha&0\\0&\beta\end{array}\right)$, we have $C^{-1}A_\mu C = A_{\alpha^{-1}\beta\mu}$, which equals $A_\mu^2$ if and only if $\alpha^{-1}\beta = 2$.
Now $C$ is a square in the group if and only if $\alpha$ and $\beta$ are both squares in the field. So in ${\rm GL}_2(F_1)$, $A_\mu$ is conjugate to its square $A_{2\mu}$ by an element $C$ that is itself a square in the group, namely $C=\left(\begin{array}{cc}1&0\\0&2\end{array}\right)$, but this is not true in ${\rm GL}_2(F_2)$.
I am not going to give details for $n>2$ for now. I think you can do it by using the fact that there is a subgroup $C_2^{n-1}$ of ${\rm GL}_n(F)$ with centralizer isomorphic to ${\rm GL}_2(F) \times {\rm GL}_{n-2}(F)$ to reduce to the case $n=2$.