**Theorem.** Let $G$ be a group. There exists a finite 3-complex $X_G$ with $\pi_1 X_G = G$ and $H_i X_G = 0$ for $i > 1$ if, and only if, $G$ is finitely presentable and has second group homology $H_2(G) = 0$.

The more interesting question to me is whether this is possible for a *finite 2-complex*, and Jens Reinhold's answer gives us the first step there. This is possible for the group $2I$ mentioned in my comment above.

**Lemma:** Let $X$ be any CW complex. Then the cokernel of the Hurewicz map $\pi_2 X \to H_2 X$ is isomorphic to $H_2(\pi_1 X)$. (Proof: attach cells to make $X$ into a $K(\pi_1 X, 1)$; doing so kills off precisely $\pi_2 X$ inside of $H_2 X$.) This is exercise 23 in Hatcher section 4.2.

**Proof of theorem.** Pick any finite presentation $P$ of $G$ and construct the presentation complex $X_P$. This is a finite complex with $\pi_1 X_P = G$, and with $H_2(X_P)$ a free abelian group $\Bbb Z^k$ on a finite number of generators. Because the cokernel of $\pi_2(X_P) \to H_2(X_P)$ is precisely $H_2(\pi_1 X_P) = H_2(G) = 0$, it follows that $\pi_2(X_P) \to H_2(X_P) = \Bbb Z^k$ is surjective. Now choose $k$ maps $\rho_i: S^2 \to X_P$ so that these give a basis of the second homology group, and set $$X_G = X_P \cup_{i=1}^k D^3,$$ attaching these 3-cells along the $\rho_i$. The resulting complex has $H_i X_G = 0$ for $i > 1$ by a Mayer-Vietoris argument, while $\pi_1 X_G = \pi_1 X_P \cong G$ by the van Kampen theorem.