The answer is negative unless you assume that $Z$ is aspherical (and even in this case I am not sure). Take Whitehead manifold $W$, which is a contractible open 3-manifold, not homeomorphic to ${\mathbb R}^3$ and let $Z$ be the 2-dimensional skeleton of a triangulation of $W$. Then $Z$ is a counter-example.
Proof.
I claim that $W$ does not admit an exhaustion by simply-connected compact submanifolds with boundary. Indeed, if such an exhaustion $W_i$ exists, then each $W_i$ is bounded by some 2-spheres, all but one of which bounded bounds a 3-ball in $W$. Filling in these spheres by 3-balls in $W$, we obtain an exhaustion $W_i'$ of $W$ by closed 3-balls. This would imply that $W$ is homeomorphic to the 3-space. Contradiction.
Your question is equivalent to asking if $Z$ (provided that it is locally finite) is exhaustable by finite simply-connected subcomplexes. Note that my $Z$ is locally finite. Suppose that such an exhaustion $Z_i$ does exist, we can then enlarge each $Z_i$ by adding to it 3-simplices in $W$ whose boundary is contained in $Z_i$. This does not change $\pi_1$, of course. The resulting subcomplexes $V_i$ will exhaust $W$ and will be simply-connected. Taking $W_i$ to be a regular neighborhood of $V_i$ we obtain an exhaustion of $W$ by simply-connected compact submanifolds with boundary. This contradicts Part 1.
