Let $G$ be a finite simple group of type ${\rm L}\_{n}^{\epsilon}(q)$ with the following conditions:  

$n$ and $\dfrac{q^{n}-\epsilon}{(q-\epsilon)(n,q-\epsilon)}$ both prime, $n\geqslant3$ and $(n,q,\epsilon)\neq(3,4,+),(3,3,-),(3,5,-),(5,2,-)$


Show that:  

1- Every minimal subgroup $L$ of order $s=\dfrac{q^{n}-\epsilon}{(q-\epsilon)(n,q-\epsilon)}$ is contained properly in a maximal subgroup $M$ isomorphic to $\mathbb{Z}\_{s}\rtimes\mathbb{Z}\_{n}$  

2- The maximal subgroup $M$ is the only proper subgroup of $G$ which contains properly $L$, i.e. every minimal subgroup of order $s$ has a unique overgroup.


As far as I checked the "ATLAS" the above propositions both are true.



Maximal subgroups of some special cases of ${\rm L}_{n}^{\epsilon}(q)$
It looks to me as though the answer is yes. You have excluded those cases where $C\_s \rtimes C\_n$ is not maximal in $L\_n^\epsilon(q)$.


So in any other situation, $C\_s \rtimes C\_n$ is maximal, and hence if $C\_s$ were contained in some other maximal subgroup $M$ then we would have $M \cap C\_s = C\_s$. But then $C\_s$ would be a self-normalizing Sylow $s$-subgroup of $M$, and so $M$ would have a normal $s$-complement by Burnside's Transfer Theorem. This rules out $M$ being almost simple, and the fact that $C\_s$ acts irreducibly implies that $M$ cannot be a geometric-type maximal.


