When I was working on this question earlier today, I made up a superficially-reasonable-sounding non-theorem in a commutative ring with no zero divisors, all irreducibles are prime. The correct theorem is: in a GCD domain, all irreducibles are prime, as mentioned by Bill Dubuque in this comment and ultimately this answer.
The fake theorem is simply not true since $\mathbb{Z}[\sqrt{-5}]$ has non-prime irreducibles (6 factors as a product of two irreducibles in two ways, as $2 \cdot 3$ and $(1 + \sqrt{-5})(1 - \sqrt{-5})$) and no zero divisors.
My original post didn't make sense, as Arturo Magidin pointed out in the comments. The thing I was trying to prove really was false and I demonstrated it with the example I chose.
So, the fake proof below is definitely fake.
However, I think it is an interesting fake. It seemed correct (or at least plausible) to me half an hour ago and I'm curious which step is wrong or unjustified.
I'm defining an integral domain as a ring with no nonzero zero divisors. There are some other equivalent definitions.
Fake Theorem: In all commutative rings with no nonzero zero divisors, all irreducible elements are prime.
Fake Proof:
Let $R$ be a ring with no nonzero zero divisors.
If $R$ is trivial it has no irreducible elements and we're done.
Let $a$ be an irreducible element of $R$ and let $(a)$ be its ideal as usual.
I will now show that $R/(a)$ is an integral domain and thus $a$ is prime.
I will start with a general product of two elements in $R/(a)$ that equals zero in $R/(a)$, expressed as representatives in $R$ of their corresponding equivalence classes, and then show that at least one of the products must be equivalent to zero in $R/(a)$. I assume a normal form property of $(c+da)$ and $(e+ha)$ where the first term is $0$ if the sum as whole is a multiple of $a$. For example $(a + 2a)$ is not in normal form, but the equivalent expression $(0 + 3a)$ is in normal form.
Suppose $(c+da)(e + ha) = ga$ where $a \nmid c$ and $a \nmid e$ (since $(c+da)$ and $(e+ha)$ are in normal form).
Since $a$ is irreducible, it also holds that $c \nmid a$ and $e \nmid a$.
$$ (c+da)(e + ha) \\ ce + eda + cha + dhaa \\ ce + (ed + ch + dha)a $$
Since $a$ is irreducible and $ce | a$ must hold in order for the product to be zero in $R/(a)$, it follows that $ce = 0$.
Thus $c$ is $0$ or $e$ is $0$, by our hypothesis that $R$ was an integral domain initially.
$ga$ factors as $(c+da)(e+ha)$, and at least one of $c$ and $e$ must be zero, so at least one of $c+da$ and $e+ha$ must be equivalent to zero in $R/(a)$.
Thus $R/(a)$ is an integral domain, since any pair of elements that multiplies to zero contains at least one element that is zero.
