Solution:
Let's suppose that $A^2 = -I_3$, then $A^2+I_3=0$
That means that $A$ is a root of the polynomial $t^2+1$.
However, since we are dealing with real values only, this polynomial has no roots, and so it is irreducible in $\mathbb R$.
But it is still a polynomial that $A$ is a root of, and so, from the definition of the minimal polynomial (let's call it $m_A$):
$m_A|t^2+1$ but $t^2+1$ is irreducible and so $m_A=t^2+1$
Invoking the Cayley-Hamilton theorem, we know that the characteristic polynomial of $A$ , $p_A$ can be divided by the minimal polynomial, and it has the same roots, so we can say that:
$p_A =(m_A)^r = (t^2+1)^r$, $r \in \mathbb N$
But since $A$ is a 3-by-3 matrix, the characteristic polynomial should be in a degree of 3:
$deg(p_A)=deg((t^2+1)^r) = 2r = 3$ so 3 is an even number. Contradiction!