Claim 1: No group

Claim 2: If $q \neq 2,3$, then the quotients of $GL(2,q)$ are exactly the subgroups of the cyclic group $\mathbb{F}^\times_q$ and the groups $GL(2,q)/D$ where $D \le \mathbb{F}^\times_q \cdot I$.

In case $GL(2,2) = S_3$, the only quotient is $\mathbb{Z}_2$.

Proof: 1) Let $q$ be odd. We have the extension

If $q = 2^n$ then $|GL(2,q)|= (q^2-1)(q^2-q) = q(q^2-1)(q-1)$. Hence a Sylow 2-subgroup $P$ is given by upper triangular matrices with unit diagonal. Thus $P \cong \mathbb{F}_q$ is an elementary abelian 2-subgroup of rank n.

2) Write $GL := GL(2,q)$, $SL := SL(2,q)$ and let $C = \langle \pm 1 \rangle$ be the center of $SL$.

Let $N$ be a normal subgroup of $GL$ and let be $H = N \cap SL$. Then $HC$ is a normal subgroup of $SL$ and $HC/C$ is a normal subgroup of the simple group $PSL = SL/C$. Hence $HC = C$ or $HC = SL$.

Case 1: $HC = C$. Hence $H \le C$. Fix $a \in N$. We want to show that $a$ is central in $GL$.

If $x \in GL$ then $xax^{-1}a^{-1} \in N \cap SL \le C$ (note: $SL$ is the commutator subgroup of $GL$). If $\operatorname{char}(\mathbb{F}_q) = 2$, then $C=1$ and we are done. So assume $\operatorname{char}(\mathbb{F}_q) > 2$. Define a map $f: GL \to C, x \mapsto [x,a]$. Since $C$ is central we have $$f(xy) = xyay^{-1}x^{-1}a^{-1}= xf(y)ax^{-1}a^{-1}=f(x)f(y)$$i.e. $f$ is a group hom. into an abelian group. Hence $SL = [GL,GL]$ lies in the kernel of $f$, i.e. $SL$ commutes with $a$. Let $\alpha \in \mathbb{F}^\times_q$ be a generator and let $x_0 = \operatorname{diag}(\alpha,1)$. A direct calculation shows that $ax_0=-x_0a$ is not possible. Therefore $a,x_0$ commute and since $GL$ is generated by $SL$ and $x_0$, we conclude that $a$ is central. Hence $N$ is a central subgroup of $GL$.

Case 2: $HC=SL$. Suppose $C \nsubseteq H$. Since $C$ is a central subgroup of $SL$ of prime order, $H \cap C = 1$ and $SL=H \times C$ follows, contradicting the indecomposability of $SL$. Thus $H=SL$ and $SL \le N$. Therefore $\mathbb{F}^\times_q = GL/SL$ maps onto $GL/N$, i.e. $GL/N$ is a quotient of $\mathbb{F}^\times_q$ and as such isomorphic to a subgroup of $\mathbb{F}^\times_q$.

2 added 54 characters in body

Claim: No group $$A_{a,b,,c} := \mathbb{Z}/2^a \times \mathbb{Z}/2^b \times \mathbb{Z}/2^c$$ with $a,b,c > 1$ is a subgroup of some $GL(2,q)$.

Proof: Let $q$ be odd. We have the extension $$1 \to SL(2,q) \to GL(2,q) \to \mathbb{F}^\times_q \to 1$$ with $\mathbb{F}^\times_q$ cyclic (as multiplicative group of finite field). Let $P$ be a Sylow 2-subgroup $GL(2,q)$. Then we have the extension $$1 \to SL(2,q) \cap P \to P \to C \to 1$$ where $C$ is a cyclic 2-group. It's known that the Sylow 2-subgroups of of $SL(2,q)$ are cyclic or generalized quaternion (see this link). Hence $GL(2,q)$ has no subgroups of the form $$A_{a,b,,c} := \mathbb{Z}/2^a \times \mathbb{Z}/2^b \times \mathbb{Z}/2^c$$ A_{a,b,,c}$with$a,b,c > 0$. If$q = 2^n$then$|GL(2,q)|= (q^2-1)(q^2-q) = q(q^2-1)(q-1)$. Hence a Sylow 2-subgroup$P$is given by upper triangular matrices with unit diagonal. Thus$P \cong \mathbb{F}_q$is an elementary abelian 2-subgroup of rank n. As a result, no group$A_{a,b,c}$with$a,b,c>1$is a subgroup of an$GL(2,q)$. 1 Let$q$be odd. We have the extension $$1 \to SL(2,q) \to GL(2,q) \to \mathbb{F}^\times_q \to 1$$ with$\mathbb{F}^\times_q$cyclic (as multiplicative group of finite field). Let$P$be a Sylow 2-subgroup$GL(2,q)$. Then we have the extension $$1 \to SL(2,q) \cap P \to P \to C \to 1$$ where$C$is a cyclic 2-group. It's known that the Sylow 2-subgroups of of$SL(2,q)$are cyclic or generalized quaternion (see this link). Hence$GL(2,q)$has no subgroups of the form $$A_{a,b,,c} := \mathbb{Z}/2^a \times \mathbb{Z}/2^b \times \mathbb{Z}/2^c$$ with$a,b,c > 0$. If$q = 2^n$then$|GL(2,q)|= (q^2-1)(q^2-q) = q(q^2-1)(q-1)$. Hence a Sylow 2-subgroup$P$is given by upper triangular matrices with unit diagonal. Thus$P \cong \mathbb{F}_q$is an elementary abelian 2-subgroup of rank n. As a result, no group$A_{a,b,c}$with$a,b,c>1$is a subgroup of an$GL(2,q)\$.