Let $x$ and $y$ be two element of general linear group $SL(n,q)$, such that the orders of $x$ and $y$ be some primitive prime divisor of $q^n-1$. Is it true that if $xy\not=yx$, then $x$ and $y$ generate $SL(n,q)$ ?
Note that the order of an element like $x$ of $SL(n,q)$ is a primitive prime divisor of $q^n-1$ if and only if $|x|\mid (q^n-1)$ but $|x|\not\mid(q^k-1)$ for every $k<n$.
Not necessarily. For $n$ even, they could generate one of the smaller classical groups ${\rm Sp}(n,q)$ or $\Omega^-(n,q)$.
For $n$ odd, it might be true sometimes - I think it is true in ${\rm SL}(5,2)$ for example. But there will sometimes be exceptions. For example, in ${\rm SL}(5,3)$ $x,y$ could generate $L_2(11)$ or $M_{11}$.
A full classification of subgroups of $GL_n(q)$ containing elements of order a primitive prime divisor of $q^n-1$ is available in the literature. See here:
Robert Guralnick, Tim Penttila, Cheryl E. Praeger, and Jan Saxl. Linear groups with orders having certain large prime divisors. Proc. London Math. Soc. (3), 78(1):167–214, 1999.
To give a full list of counter-examples to the OP's question, then, one needs only find which of the groups mentioned in this paper are non-abelian and lie in $SL_n(q)$.
In addition to the classical groups that Derek mentions the cited paper lists a number of `geometric' subgroups that satisfy the required conditions. These include the field-extension subgroups (i.e. Aschbacher's $\mathcal{C}_3$ class) whenever $n$ is a composite. So, for instance $SL_6(q)$ contains $SL_3(q^2)$ and $SL_2(q^3)$, both of which are non-abelian and have primitive prime divisors of $q^6-1$ for $q>2$.
Apart from these, all counter-examples are `nearly simple' i.e. their projective image is an almost simple group. These include the sporadic examples mentioned by Derek plus a bunch of others.
In fact the paper cited above doesn't just deal with ppd's of $q^n-1$ but ppd's of $q^e-1$ for $e>\frac{n}{2}$.
