There are good reasons behind the convention of including $(0)$ as a prime ideal
but excluding $(1).\ $ First, we include zero as a prime ideal because it facilitates many useful reductions. For example, in many ring theoretic problems involving an ideal $\, I\,$, one can reduce to the case $\,I = P\,$ prime, then reduce to $\,R/P,\,$ thus reducing to the case when the ring is a domain. In this case one simply says that we can **factor out by the prime** $ P\,$, so w.l.o.g. assume $\, P = 0\,$ is prime, so $\,R\,$ is a domain. For example, I've appended to the end of this post an excerpt from Kaplansky's classic textbook *Commutative Rings*, section $1\!\!-\!\!3\!:\,G$-Ideals, Hilbert Rings, and the Nullstellensatz.

Thus we have solid evidence for the utility of the convention that the zero ideal is prime. So why don't we adopt the same convention for the unit ideal $(1)$ or, equivalently, why don't we permit the zero ring as a domain? There are a number of reasons. First, in domains and fields it often proves very convenient to assume that one has a nonzero element available. This permits proofs by contradiction to conclude by deducing $\,1 = 0.\ $ More importantly, it implies that the unit group is nonempty, so unit groups always exist. It'd be very inconvenient to have to always add the proviso (except if $\, R = 0)\,$ to the many arguments involving units and unit groups. For a more general perspective it's worth emphasizing that the usual rules for equational logic are not complete for *empty structures* so that is why groups and other algebraic structures are always axiomatized to prevent nonempty structures (see this thread for details).

Below is the promised Kaplansky excerpt on reduction to domains by factoring out prime ideals. I've explicitly emphasized the reductions e.g. **reduce to...**.

Let $\, I\,$ be any ideal in a ring $\, R.\,$ We write $\, R^{*}\,$ for the quotient ring $\, R/I.\,$ In the polynomial ring $\, R[x]\,$ there is a smallest extension $\, IR[x]\,$ of $\, I.\,$ The quotient ring $\, R[x]/IR[x]\,$ is in a natural way isomorphic to $\, R^*[x].\,$ In treating many problems, we can in this way **reduce to the case** $\, I = 0,\,$
and we shall often do so.

**THEOREM $28$.** $\,$ Let $\, M\,$ be a maximal ideal in $\, R[x]\,$ and suppose that the contraction $\, M \cap R = N\,$ is maximal in $\, R.\ $ Then $\, M\,$ can be generated by $\, N\,$ and one more element $\, f.\ $ We can select $\, f\,$ to be a monic polynomial which maps $\!\bmod N\,$ into an irreducible polynomial over the field $\, R/N.\ $

**Proof.** $\,$ We can **reduce to the case** $\, N = 0,\,$ i. e., $\, R\,$ a field, and then
the statement is immediate.

**THEOREM $31$.** $\,$ A commutative ring $\, R\,\,$ is a Hilbert ring if and only if the polynomial ring $\, R[x] \,\,$ is a Hilbert ring.

**Proof.** $\,$ If $\, R[x]\,$ is a Hilbert ring, so is its homomorphic image $\, R\,$.
Conversely, assume that $\, R\,$ is a Hilbert ring. Take a G-ideal $\, Q\,$ in
$\, R[x]\,$; we must prove that $\, Q\,$ is maximal. Let $\, P = Q \cap R\,$; we can **reduce the problem to the case** $\, P = 0,\,$ which, incidentally, makes $\, R\,$ a domain.
Let $\, u\,$ be the image of $\, x\,$ in the natural homomorphism $\, R[x] \to R[x]/Q.\,$
Then $\, R[u]\,$ is a G-domain. By Theorem $23$, $\,u\,$ is algebraic over $\,R\,$ and $\,R\,$ is a G-domain. Since $\,R\,$ is both a G-domain and a Hilbert ring, $\,R\,$ is a field. But this makes $\, R[u] = R[x]/Q\,$ a field, proving $\, Q\,$ to be maximal.