$$\newcommand{\crhom}[1]{\mathbf{CRing}(\mathbb{Z}[X], #1)}$$

Theorem: The forgetful functor $U : \mathbf{CRing} \rightarrow \mathbf{Set}$ is representable. In particular $U \cong \crhom{\dash}$.

Proof: First, we establish the following fact. For a commutative ring $R$ and element $r \in R$, define the evaluation map $e_r : \mathbb{Z}[X] \rightarrow R$ by $e(p(X)) = p(r)$, the map that evaluates the given polynomial at $r$ inside $R$. Note that $e_r(X) = r$. This is indeed the unique homomorphism $\mathbb{Z}[X] \rightarrow R$ that sends $X \mapsto r$. This is straightforward to see: Suppose $f : \mathbb{Z}[X] \rightarrow R$ were a homomorphism with $f(X) = r$. Let $p(X) = \sum_0^kp_n X^n$. Then,

$$f(p(X)) = f\left( \sum_0^k p_n X^n \right) = \sum_0^k p_n f(X)^n = \sum_0^k p_n r^n = e_r(p(X))$$

and thus $f = e_r$.

The preceding fact is most of the content of the proof. Indeed, define the natural transformation $\alpha : U \rightarrow \crhom{\dash}$ with components $\alpha_R : U(R) \rightarrow \crhom{R}$ taking an element $r$ to the map $e_r$. We verify naturality: Let $f : R \rightarrow S$ be a ring homomorphism. We must have $\alpha_S \circ U(f) = \crhom{f} \circ \alpha_R$. Again $\crhom{f}$ is just left-composition by $f$, and $U(f)$ is just $f$. So, let $r \in U(r)$. Applying the right side to $r$ yields $f \circ e_r$. Applying the left side to $r$ yields $e_{f(r)}$. But of course,

$$e_{f(r)}\left( \sum_0^k p_n X^n \right) = \sum_0^k p_n f(r)^n = f \left( \sum_0^k p_n r^n \right) = f \left( e_r \left( \sum_0^k p_n X^n \right) \right)$$

and $e_{f(r)} = f \circ e_r$. So $\alpha$ is natural and $U \cong \crhom{\dash}$.