Suppose $G$ is a finitely generated subgroup of $GL_n(Z)$, $n\ge 3$. I suspect that there is no decision procedure for deciding whether or not such $G$ is finitely presented. How can this proved?
Various forms of this question were discussed on MO several times, see e.g. here. The key is that $SL(4,Z)$ contains $H=F_2\times F_2$, direct product of rank 2 free groups. Then for every finitely-presented group $Q$ (i.e., a group $Q$ with a given finite presentation), there is an epimorphism $H\to Q$ with finitely-generated kernel $K$, see e.g. here; this is called Mikhailova's construction. Unless $Q$ is finite, $K$ is not finitely-presented, see the same link. Now, take a finitely-presented group $Q$ for which it is undecidable if it is finite or not. Then for $K$ it will be undecidable if it is fp or not.
Situation is different for $GL(3,Z)$ as it contains no direct products of free nonabelian groups. It is an open problem due to J.-P. Serre (1979 or so, depending how you count), if $SL(3,Z)$ is coherent (i.e., if every finitely-generated subgroup of $SL(3,Z)$ is also finitely-presentable). Situation with $GL(3,Z)$ is the same, of course. At this moment, all known (say, torsion-free) f.g. subgroups of $SL(3,Z)$ fall into the following classes:
a. Finite index. b. Closed surface subgroups. c. Free subgroups. d. Non-Zariski dense in $SL(3,R)$ subgroups (abelian subgroups and semidirect products of free groups of rank $1\le r< \infty$ with $Z^2$).
All of such subgroups are finitely-presented, of course, so, for all what we know, $SL(3,Z)$ is coherent, in which case your question has obvious answer. If $SL(3,Z)$ is not coherent, who knows...