Those are the only homomorphisms. (Except, if you don't require ring homomorphisms to preserve $1$, for the constant zero morphism).
Let $e_i$ be the sequence whose $i$th element is $1$ and the rest is $0$.
If $f:\mathbb Z^{\mathbb N}\to\mathbb Z$ is a ring homomorphism, then at most one of $f(e_0), f(e_1), \ldots, f(e_n), \ldots$ can be nonzero. Namely, if $i\ne j$, then $e_ie_j=0$ and therefore $f(e_ie_j)=f(e_i)f(e_j)=0$, so one of $f(e_i)$ and $f(e_j)$ must be zero.
Thus $f$ agrees with some multiple of a projection on the ideal generated by the $e_i$s. Without loss of generality, let's assume that $f(x_0,x_1,\ldots)=ax_0$ for some $a$ whenever cofinitely many $x_i$s are zero.
The following argument is adapted from an MO post that Mike Miller pointed to in a comment.
Choose sequences $(\alpha_i)_i$, $(\beta_i)_i$ such that $2^i\mid \alpha_i$ and $3^i\mid \beta_i$ and $\alpha_i+\beta_i=1$. (This is possible due to Bézout's identity since $2^i$ and $3^i$ are always coprime).
Now to show $f(x_0,x_1,x_2,\ldots)=ax_0$ in general write
$$ f(x_0,x_1,x_2,x_3,\ldots_)=
\underbrace{f(x_0,0,0,\ldots)}_{ax_0}
+\underbrace{f(0,\alpha_1 x_1,\alpha_2 x_2,\alpha_3 x_3,\ldots)}_P
+\underbrace{f(0,\beta_1 x_1, \beta_2 x_2, \beta_3 x_3, \ldots)}_Q $$
I claim that each of the two last terms is zero. For any $n$ we have
$$ P
= \underbrace{f(0,\alpha_1 x_1, \ldots, \alpha_{n-1} x_{n-1}, 0, 0, \ldots)}_0
+ \underbrace{f(0,0,\ldots,0,\alpha_n x_n,\alpha_{n+1} x_{n+1}, \ldots)}
_{\text{divisible by }2^n} $$
The only integer that is divisible by $2^n$ for all $n$ is zero, so $P$ must be $0$. Similarly $Q$ must be $0$.
So $f(x_0,\ldots)=ax_0$, and since $f(1,1,1,\ldots)=1$ (otherwise $f$ is not a ring homomorphism) we get $a=1$ and $f$ is a projection.
Bonus answer: This argument works whenever $R$ is a UFD with at least two different prime elements.
