If $n>2$, this is a particular case of the main result in [G. Prasad, Discrete subgroups isomorphic to lattices in semisimple Lie groups, Amer. J. Math. 98 (1976), no. 1, 241--261], namely irreducible lattices in linear semisimple Lie groups are co-Hopf (where a group is called * co-Hopf* if it is not isomorphic to its proper subgroup). I think, $GL(2,\mathbb Z)$ is also co-Hopf, but at the moment I am not sure how to prove this; note that $GL(2,\mathbb Z)$ has $\mathbb Z*\mathbb Z$ as a finite index subgroup, and $\mathbb Z*\mathbb Z$ isn't co-Hopf.

EDIT: As I mentioned in comments, Prasad's paper implies the result when the ambient Lie group is semisimple, which covers the cases of $SL(n,\mathbb Z)$ and $PGL(n,\mathbb Z)=PSL(n,\mathbb Z)$ when $n>2$. I cannot find a cheap proof for $GL(n,\mathbb Z)$, but here is an ad hoc argument.

A key point is that any injective endomorphism $\phi$ of $GL(n,\mathbb Z)$ must have finite cokernel (i.e. its image has finite index). Indeed, its restriction to $SL(n,\mathbb Z)$ followed by projection $GL(n,\mathbb Z)\to PGL(n,\mathbb Z)$ is a homomorphism of lattices $\phi_0: SL(n,\mathbb Z)\to PSL(n,\mathbb Z)$ in locally isomorphic semisimple Lie groups so Margulis superrigidity implies that $\phi_0$ has finite cokernel, and hence so does $\phi$. Now if $-I_n$ is not in the image of $\phi$, then $GL(n,\mathbb Z)$ embedds as a finite index subgroup into $PGL(n,\mathbb Z)$, so
by $GL(n,\mathbb Z)$ is isomorphic to a lattice in a semisimple Lie group, hence it is co-Hopf by Prasad. If $-I_n$ lies in the image of $\phi$, then $-I_n=\phi(-I_n)$, so $\phi$ descends to an injective endomorphism of $PGL(n,\mathbb Z)$, which by Prasad is onto, and it easily implies that $\phi$ is onto.