What other properties follow from having a ring homomorphism to $\mathbb{Z}$? (All my rings have $1$, and ring homomorphisms preserve $1$.)
In $\mathbf{Set},$ the points of an object $X$ can be thought of as arrows from the terminal object $1$ to $X$. So I guess in general, we could say that a "copoint" of an object $X$ is an arrow to the initial object of whatever category $X$ belongs to.
In $\mathbf{Set}$, this is pretty boring, because a set has a copoint iff it is empty. Its also pretty boring in categories like $\mathbf{Ab}$ with a $0$ object, for a different reason: every object has precisely one copoint in this case.
In $\mathbf{Ring},$ things are more interesting. If $R$ is a ring, then a copoint of $R$ is a ring homomorphism $f:R \rightarrow \mathbb{Z}$. Hence the existence of a copoint implies that $R$ has characteristic $0$, since for any integer $k \in \mathbb{Z}$, if we assume $k1_R = 0_R$, then $f(k1_R) = 0_\mathbb{Z}$, so $kf(1_R) = 0_\mathbb{Z}$, so $k1_\mathbb{Z} = 0_\mathbb{Z}$, so $k=0$.

Question. What other properties follow from having a ring homomorphism to $\mathbb{Z}$?

 A: Suppose we have a ring homomorphism $f\colon R\to\mathbb{Z}$.
The composite of the homomorphism $f$ following the homomorphism $j\colon \mathbb{Z}\to R : n\mapsto n\cdot 1_R$ is
the identity homomorphism $f\circ j=\mathrm{id}_{\mathbb{Z}}$ (this because $\mathrm{id}_{\mathbb{Z}}$ is the only endomorphism of the ring $\mathbb{Z}$), which implies that $j$ is injective, and hence the characteristic of $R$ is zero. Note that $\mathrm{char}(R)=0$
if and only if the element $1_R$ of the $\mathbb{Z}$-module $R$ is torsion-free,
if and only if the set $\{1_R\}$ is linearly independent over $\mathbb{Z}$.
This was just an alternative proof of what you have already proved.
Let $N$ be the kernel of $f$. Then the underlying additive group of the ring $R$ is the (inner) direct sum of the additive groups of the subring $\mathbb{Z}1_R$ and of the ideal $N$, which we write $R=\mathbb{Z}1_R\oplus N$. Indeed, if $x$ is any element of the ring $R$, then $f(x-f(x)1_R)=0$ and therefore $x-f(x)1_R\in N$, which proves that $R=\mathbb{Z}1_R+N$. If $x=n\cdot 1_R\in N$ (where $n\in\mathbb{Z}$), then $n=f(x)=0$ and hence $x=0$, which proves that $\mathbb{Z}\cap N=0$, so the sum $\mathbb{Z}1_R+N$ is direct.
The ideal $N$, equipped with the addition and multiplication inherited from $R$, is a rng. $\,$(This "rng" is not a typo: a rng is an abelian group equipped with an associative biadditive multiplication. Nothing is said of a multiplicative identity; even if it is present, it is ignored in the sense that homomorphisms of rngs are not required to preserve multiplicative identities, since a multiplicative identity is not an officially acknowledged constituent of a rng structure.)$\,$ The inclusion map $N\hookrightarrow R$ is a homomorphism or rngs,
where the ring $R$ is regarded as a rng (we 'forget' $1_R$).
The "ring regarded as a rng" is in fact a forgetful functor $U$ from the category $\mathbf{Ring}$ of (unital) rings to the category $\mathbf{Rng}$ of rngs. The left adjoint $F\colon\mathbf{Rng}\to\mathbf{Ring}$ constructs for each rng $A$ a ring $F\mspace{-2mu}A$ and a homomorphism of rngs $\eta=\eta_A\colon A\to UF\mspace{-2mu}A$ so that the following is true: given any
ring $R$ and a homomorphism of rngs $h\colon A\to UR$ there exists a unique homomorphism of rings $h'\colon F\mspace{-2mu}A\to R$ such that $h=Uh'\circ\eta$. $\,$(Here the homomorphism of rngs $Uh'\colon UF\mspace{-2mu}A\to UR$ is precisely the same mapping as the homomorphism or rings $h'$;
though the mapping $h'$ sends the multiplicative identity of the ring $F\mspace{-2mu}A$ to the multiplicative identity of the ring $R$, we ignore this fact when we regard $h'$ as a homomorphism of rngs and write it $Uh'$.)$\,$ I am sure you know the construction of $F\mspace{-2mu}A$. We set $F\mspace{-2mu}A=\mathbb{Z}\oplus\mspace{-2mu}A$, write elements of the direct sum $\mathbb{Z}\oplus\mspace{-2mu}A$ as formal sums $n+a$ ($n\in\mathbb{Z}$, $a\in A$), which we add componentwise and multiply them as follows: $(n+a)(m+b):=(nm)+(nb+ma+ab)$. The homomorphism $\eta\colon A\to F\mspace{-2mu}A$ of rngs maps an element $a$ of $A$ to the formal sum $0+a$ in $F\mspace{-2mu}A$. The homomorphism $\eta$ embeds the rng $A$ into the ring $F\mspace{-2mu}A$ as the subrng $\eta A=0\oplus\mspace{-2mu} A$. $\,$And now comes the punchline of this particular joke:
the mapping $\varphi\colon F\mspace{-2mu}A\to\mathbb{Z} : n+a\mapsto n$ is a homomorphism of rings, and $\ker\varphi=0\oplus\mspace{-2mu} A=\eta A$.
In the situation with which we began, the ring $FN$ is naturally isomorphic to the ring $R$: the natural isomorphism $FN\to R$ maps a formal sum $n+z\in FN$ ($n\in\mathbb{Z}$, $z\in N$) to the element $n\cdot 1_R+z$ of the ring $R$.
Therefore, the special property of a ring $R$, besides $\mathrm{char}(R)=0$, which actually characterizes the existence of a ring homomorphism $R\to\mathbb{Z}$, is the existence of a split $R=\mathbb{Z}1_R\oplus N$, where $N$ is
a two-sided ideal of $R$ and the direct sum is that of the underlying additive groups.
For every rng $A$ there exists a ring homomorphism $R\to\mathbb{Z}$ whose kernel is rng-isomorphic to $A$.
In a certain sense a ring homomorphism $f\colon R\to\mathbb{Z}$ is just a 'ringed' presentation of the rng $N=\ker F$. Another way to describe this correspondence is by saying that the notion of a ring homomorphism $R\to\mathbb{Z}$ is cryptomorphic to the notion of a rng; the two notions are just two equivalent presentations/realizations/implementations/incarnations of the same concept.
Let $\mathfrak{R}$ be the category or ring homomorphisms $R\to\mathbb{Z}$, where a homomorphism of $f\colon R\to\mathbb{Z}$
to $g\colon S\to\mathbb{Z}$ is a homomorphism of rings $h\colon R\to S$ such that $f=g\circ h$. $\,($This category $\mathfrak{R}$ is
the comma category $(\mathbf{Rings}\downarrow\mathbb{Z})$.$)\,$ The category $\mathfrak{R}$ is equivalent to the category of rngs $\mathbf{Rng}\,$;
figure out on your own the pair of functors that constitute the equivalence of these two categories -- you have almost all ingredients already at hand. This equivalence of categories can be regarded as the 'cryptomorphism' between the notion of a ring homomorphism $R\to\mathbb{Z}$ and the notion of a rng.
