Let $q$ be a prime power and $n=lm$ an integer with $l,m>1$. We know that the finite field $GF(q^n)$ is a $n$-th dimensional vector space over $GF(q)$, and it is also a $l$-th dimensional vector space over $GF(q^m)$.
Suppose that $S$ is a set of elements of $GF(q^n)$ which is linearly independent over $GF(q)$. Given that $S$ does not contain any elements of $GF(q^m)$, can we say something about the linear dependence or independence of $S$ over $GF(q^m)$? Are there any conditions on $q,m,n,l$ that would make $S$ linearly independent over $GF(q^m)$ too?
You seem to be asking too much. Pick an element $\alpha\in GF(q^m)\setminus GF(q)$ and an element $\beta\in GF(q^n)\setminus GF(q^m)$. Then $\beta$ and $\beta\alpha$ are obviously $\notin GF(q^m)$. They are linearly independent over $GF(q)$ because their ratio $\alpha\notin GF(q)$. Yet they are clearly linearly dependent over $GF(q^m)$.
A related concept that may or may not help you is that of linearly disjoint field extensions. Given a field $K$ we say that two extension fields, $L$ and $L'$ (both inside a common bigger field), are linearly disjoint over $K$, if any subset of $L$ that is linearly independent over $K$ remains linearly independent over $L'$ (or equivalently any subset of $L'$ that is linearly independent over $K$ remains linearly independent over $L$.
For $L$ and $L'$ to be linearly disjoint over $K$ it is clearly necessary that $L\cap L'=K$. It can be shown that if $L/K$ and $L'/K$ are both Galois, then this necessary condition is also sufficient. In your case this implies that if (and only if) $\gcd(\ell,m)=1$, then the fields $GF(q^m)$ and $GF(q^\ell)$ are linearly disjoint over $GF(q)$. So in that case elements of $GF(q^m)$ that are inearly independent over $GF(q)$ remain so over $GF(q^\ell)$.
Our study group was once in dire need of basic information about this topic, so I wrote a quick and dirty intro myself.