This never happens for finite fields $F \neq \mathbb{F}_2$.  If a group $G$ acts on an abelian group $M$, then short exact sequences

$1 \rightarrow M \rightarrow \Gamma \rightarrow G \rightarrow 1$

are classified by elements of $H^2(G;M)$.  It is thus enough to show that if $F \neq \mathbb{F}_2$ is a finite field and $V = F^n$, then $H^2(GL_n(F);V)=0$.  In fact, we will show that $H^k(GL_n(F);V)=0$ for all $k$.

We have a short exact sequence

$1 \rightarrow F^{\times} \rightarrow GL_n(F) \rightarrow PGL_n(F) \rightarrow 1.$

Associated to this is the Hochschild-Serre spectral sequence in cohomology with coefficients in $V$.  The $E_2$-term is $H^p(PGL_n(F);H^q(F^{\times};V))$.  The key fact here is that $H^q(F^{\times};V)=0$ for all $q$.

On page 58 of Brown's book on group cohomology, there is a calculation of the cohomology of finite cyclic groups with nontrivial coefficients.  In the case we're considering, it goes as follows.  Define $N = \sum_{x \in F^{\times}} x \in \mathbb{Z}[F^{\times}]$ (of course, $N$ acts as $0$ on $V$, but forget that for the moment).  We then get a map $N : V \rightarrow V$ whose image lies in the ring of invariants $V^{F^{\times}}$ and which satisfies $N(gv)=N(v)$ for all $g \in F^{\times}$ and all $v \in V$.  Let $V_{F^{\times}}$ be the ring of coinvariants, ie the quotient of $V$ by the subspace spanned by $\langle g v-v\ |\ g \in F^{\times},\ v \in V\rangle$.  We get an induced map $\overline{N}:V_{F^{\times}} \rightarrow V^{F^{\times}}$.  The result then is that $H^0(F^{\times};V) = V^{F^{\times}}$, that $H^i(F^{\times};V) = ker\ \overline{N}$ for $i \geq 1$ odd, and that $H^i(F^{\times};V) = coker\ \overline{N}$ for $i \geq 1$ even.  But clearly $V^{F^{\times}} = 0$, and since $F$ is not the field with $2$ elements we also have $V_{F^{\times}} = 0$.  The result follows.