Here is a non nilpotent example.

Take $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 1 & -1 \\ 3 & 2 & 0 \end{pmatrix}$. Then $A$ is not special, but $A^2 = \begin{pmatrix} 8 & 5 & 1 \\ 0 & 0 & 0 \\ 8 & 5 & 1 \end{pmatrix}$ is. Moreover, $A^{r+2} = 3^r\cdot A^2$ for any $r\geq 1$, so $A$ is not nilpotent.

Note that for $2\times 2$ matrices, if a power of a matrix is special, then the matrix itself is. Indeed, assume that the two rows of $A^k$ are equal and non-zero. Since any special matrix is singular, then there is a scalar $\lambda$ such that the second row of $A$ is $\lambda$ times the first. This is still true for any power of $A$; thus the second row of $A^k$ is $\lambda$ times the first. Since they are equal and non-zero, $\lambda = 1$, so $A$ is special.

**Added to fit the comments to the question:**

If you assume that only one of the eigenvalues of the $n\times n$ matrix $A$ is $0$, then the answer becomes negative. Indeed, in that case, $A^r$ has rank $n-1$ for any $r\geq 1$. Without loss of generality, we can assume that the first $n-1$ rows of $A$, say $R_1, \ldots, R_{n-1}$, are linearly independent. The last row is then some linear combination $R_n = L(R_1, \ldots, R_{n-1})$ of the others.

But then, if $R_1^{(r)}, \ldots, R_n^{(r)}$ are the rows of $A^r$, we still have that $R_n^{(r)}=L(R_1^{(r)}, \ldots, R_{n-1}^{(r)})$, so the first $n-1$ rows of $A^r$ are still linearly independent (since $A^r$ has rank $n-1$). If $A^r$ is special, then $R^{(r)}_n$ has to be equal to some $R_i^{(r)}$; this means that the coefficients in $L$ are all zero, except the one in front of $R_i^{(r)}$, which is $1$. Thus $R_n = R_i$, and $A$ is special.

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