I am aware of two obvious examples of matrices which are their own square:
Are there any other matrices such that $A^2$ = $A$?
My original line of reasoning was to multiply each side by $A^{-1}$ to get the result:
$A^{-1}A^2 = A^{-1}A$
$A = I$
However, this is clearly wrong because the 0 matrix satisfies the conditions and is not an identity matrix.
On
There are quite a few examples, a simple example has been given by Dave in the comment \begin{bmatrix} 0&0\\0&1 \end{bmatrix} and \begin{bmatrix} 1&0\\0&0 \end{bmatrix} I think your problem mainly come from your assumption that the matrix A is invertible, you have proved that the only invertible matrix that is its own square is the identity. However, the zero matrices are not invertible. So the square of zero matrix is also zero doesn't contradict with your previous conclusion.
\begin{eqnarray*} A= \begin{bmatrix} \frac{1 \pm \sqrt{1-4bc}}{2} & b \\ c & \frac{1 \mp \sqrt{1-4bc}}{2} \\ \end{bmatrix} . \end{eqnarray*}