I combine user37314's answer and my comments; the claim is that any smooth projective complex algebraic surface with $\pi_1$ abelian and $\pi_2$ finite has a finite cover which must be an abelian surface; in particular, $X$ cannot be of general type.

By replacing $X$ by a finite cover, we may assume that $\pi_1(X)$ is free abelian.
The map $X \to Alb(X)$ induces an isomomorphism on $\pi_1$ and the universal cover of $Alb(X)$ is a complex Euclidean space, so by taking the fibre product we get a proper map from the universal cover $\tilde{X}$ of $X$ to a complex Euclidean space. The fundamental class of any positive dimensional fibre of this map would give a non-torsion class in $H_2(\tilde{X}) = \pi_2(\tilde{X})$ (since $\pi_1$ is trivial) so this map must be finite.

It follows that $\tilde{X}$ is Stein and moreover $\pi_2(\tilde{X}) = 0$ (see Gurjar, "Two remarks on the topology of projective surfaces." Math. Ann. 328 (2004), no. 4, 701–706; we do not need any results about the Shafarevich conjecture.) Since $\tilde{X}$ is a Stein surface it follows (from Morse theory) that $H_i(\tilde{X}) = 0$ for all $i > 2$ and so by Hurewicz, $\pi_i(\tilde{X}) = 0$ for all $i$. It follows that $\tilde{X}$ is contractible, so the cohomology of $X$ must be the group cohomology of $\pi_1(X)$. Since $X$ is a projective surface this can only happen if $\pi_1(X)$ has rank $4$ and the cohomology ring of $X$ must be isomorphic (by pullback) to that of its Albanese. This implies that the Albanese map has degree $1$ and all fibres are finite so it must be an isomorphism.