Is every prime element of a commutative ring "veryprime"?

Let $p$ be an element with $p\mid ab\implies p\mid a\lor p\mid b$ and that is not a divisor of zero.

Assume $p\|ab\ne p\|a+p\|b$. Then certainly $p\|a$ and $p\|b$ are both finite. So write $a=p^ka'$, $b=p^mb'$ with $k,m\in\mathbb N_0$ and $p\nmid a'$, $p\nmid b'$. Then $ab=p^{k+m}a'b'$ and $p\|ab\ne k+m$ means that $p^{k+m+1} \mid ab$, say $ab=p^{k+m+1}c$. Then $$p^{k+m}(pc-a'b')=0. $$ As $p$ is not a divisor of zero, we conclude $pc=a'b'$, i.e., $p\mid a'b'$ contradicting $p\nmid a'$, $p\nmid b'$. We conclude that $p\|ab= p\|a+p\|b$.

Now that I've gotten this far in writing this up, I see that quid presented the counterexaple suggested by this finding. At least this elaborates on quid's closing remark. :)

The ring $\mathbb{Z}/4\mathbb{Z} $ provides a counterexample. The element $2$ in that ring is prime, and $2 \mid 2$ while $2^2 \nmid 2$ yet $2^n \mid 2^2 = 0$ for every $n$.

If one excludes zero-divisors, things should work by an inductive argument; it just needs to be possible to cancel by prime elements.