Let $\lambda$ be an eigenvalue of $A$. Then $\lambda^4$ is an eigenvalue of $A^4$.
Let $v$ be the corresponding eigenvector of $\lambda^4$. Then $A^4 v = \lambda^4 v \Rightarrow Iv=\lambda^4 v \Rightarrow \lambda^4=1 \Rightarrow \lambda= \pm 1,\pm i$.
Since $A$ is a real matrix, if $i$ (or $-i$) is it's eigenvalue, then $-i$ (respectively $i$) is also it's eigenvalue. Also $A \neq \pm I$.
Therefore possible characteristic polynomials of $A$ are $(x-1)(x+1)^2,(x+1)(x-1)^2,(x-1)(x^2+1),(x+1)(x^2+1)$.
Now if possible, $x^2+1$ will be a minimal polynomial of $A$ for the characteristic polynomials $(x-1)(x^2+1),(x+1)(x^2+1)$ only. But any such minimal polynomial must have factors $(x-1)$ and $(x+1)$ respectively.
Hence $A^2+ I=0$ is not possible.
Are there any mistakes in my proof? I am learning to handle proofs containing characteristic polynomials, minimal polynomials, eigenvalues etc. Thanks.