For a positive-size complex square matrix, nilpotence is equivalent to having only the eigenvalue zero. Nilpotence forces $\lambda^k=0$ for any eigenvalue. Conversely, if the spectrum is $\{0\}$, the characteristic polynomial is $t^n$, so Cayley–Hamilton gives $A^n=0$.
For real matrices, interpret the spectrum over $\mathbb C$. Having zero as the only real eigenvalue is not enough, since nonreal eigenvalues could also occur.
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