Patrick's answer shows that Noetherianness is *sufficient*, and I hope to show it's *necessary* by providing some counterexamples.

Clearly (ii) always implies (i) and (iii) even without the Noetherian condition.

## Example 1: Local, Krull dimension $0$, but $m^k\neq \{0\}$ for all $k$.

Take a field $F$ and the polynomial ring $F[x_2,x_3,x_4\ldots]$ in countably many variables, and take the quotient by the ideal $(x_2^2,x_3^3,x_4^4,\ldots)$. As you can see $m=(x_2,x_3,x_4,\ldots)$ is not finitely generated and nil, and it contains elements of arbitrarily high nilpotency index, so it is not nilpotent.

## Example 2: Local and $\{0\}\neq m=m^2$ .

In the same picture as above, throw in more generators to the ideal you are quotienting by to include things of the form $x_i -x_{i+1}x_{i+2}$. When this is done, $m=\{x_2,x_3,x_4,\ldots\}$ is now idempotent in addition to being the unique maximal ideal.

(Yeah, I see that example two actually does both things, but the thing is that I came up with the first one and then tinkered a long time before settling on the second one :) )