There is a cofibrant way of showing this, i.e. forgetting about coverings, Postnikov towers, fibrations, spectral sequences, etc. There are very nice and simple algebraic models for low-dimensional homotopy types. The simplest are **crossed modules**, which are group homomorphisms $$\partial\colon C_2\longrightarrow C_1$$ such that $C_1$ acts on the right of $C_2$ and the following two equations are satisfied:

$$\partial(x_2^{x_1})=x_1^{-1}x_2x_2, \qquad x_2^{\partial(y_2)}=y_2^{-1}x_2y_2.$$

Crossed modules can be regarded as non-abelian chain complexes $C_*$ concentrated in degrees $1$ and $2$. The subscript indicates the degree of each element. Notice that the first equation says that $\partial$ is $C_1$-equivariant if we let $C_1$ act on itself by conjugation.

The homology of $C_*$ is usually regarded as homotopy groups:
$$\pi_1C_*=C_1/\partial(C_2),\qquad \pi_2C_*=\ker\partial.$$
Notice that $\pi_1C_*$ acts on the right of $\pi_2C_*$.

The canonical example of a crossed module is the homomorphism
$$\partial \pi_2(X,Y)\longrightarrow \pi_1Y$$
associated to any pair of spaces $(X,Y)$. The **fundamental crossed module** of a connected CW-complex $X$ with $1$-skeleton $X^1$ is
$$\partial\colon\pi_2(X,X^1)\longrightarrow \pi_1X^1.$$

To any crossed module $C_*$ we can associate a two-step chain complex
$$\cdots\rightarrow 0\rightarrow C_2^{ab}\otimes_{\mathbb{Z}[C_1]}\mathbb{Z}\stackrel{\bar{\partial}}\longrightarrow C_1^{ab}\rightarrow 0\rightarrow \cdots$$
by abelianizing $C_1$ and $C_2$ and killing the action of $C_1$ on $C_2^{ab}$. If $C_*$ is the fundamental crossed module of $X$ then the homology of this chain complex is $H_1(X)$ and $H_2(X)$ in the corresponding degrees.

Now assume $\pi_1(X)\cong\mathbb{Z}$. Then the natural projection $C_1=\pi_1X^1\twoheadrightarrow \pi_1X\cong\mathbb{Z}$ has a section $i\colon \pi_1X\rightarrow \pi_1X^1$. This section gives rise to a homotopy equivalence of crossed modules:
$$\begin{array}{rcccl}
&\pi_2X&\stackrel{0}\longrightarrow&\pi_1X&\\\
{\text{inclusion}}&\downarrow&&\downarrow&\scriptstyle i\\\
&\pi_2(X,X^1)&\longrightarrow&\pi_1X^1&
\end{array}$$
In particular, the chain complexes associated to these two crossed modules are quasi-isomorphic. The chain complex of the upper crossed module, given by the trivial homomorphism $0\colon \pi_2X\rightarrow \pi_1X$, is simply
$$\cdots\rightarrow 0\rightarrow \pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}\stackrel{0}\longrightarrow (\pi_1X)^{ab}\rightarrow 0\rightarrow \cdots,$$
hence we recover the well-known isomorphism $(\pi_1X)^{ab}=H_1X$ and what we wanted to obtain $\pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}=H_2X$.