possible eigenvalues of $A$

I think you should consider the fact that corresponding eigenvectors of $A$ and $A^2$ are the same. So these $\lambda$ and $\lambda^2$ are both eigenvalues of the same eigenvectors. Since eigenvalues must be unique, then $\lambda = \lambda^2$, which leads $\lambda = 0$ or $\lambda = 1$.

EDIT:

Suppose $\lambda$ is a real eigenvalue of $A$, with eigenvector $v$; then $Av=\lambda v$ and, easily, also $A^2v=\lambda^2v$. Therefore $$ A^tv=A^2v=\lambda^2v $$ and so $v^tA=\lambda^2v^t$. Hence $$ v^tAv=\lambda^2(v^tv) $$ but, on the other hand, $$ v^tAv=v^t(\lambda v)=\lambda(v^tv) $$ Since $v^tv\ne0$, we get $\lambda^2=\lambda$.

If $\lambda$ is an eigenvalue, so is $\lambda^2$, and so is, for the same reason, $\lambda^{4}$, $\lambda^{8}$... or any $\lambda^{2^k}$.

Thus, in the case $\lambda$ is not in $\{-1,0,1\}$, it would generate an infinite spectrum, which cannot be.

It remains the case $\lambda=-1$ that has to be eliminated, because all eigenvalues of $A$ are eigenvalues of $A^2$, and these are $\geq 0$.