$\lambda$ is an eigenvalue of $A$ if and only if $\lambda + \mu$ is an eigenvalue of $A+\mu I$. To see this, note that $A\mathbf{v}=\lambda\mathbf{v}$ if and only if $A\mathbf{v}+\mu\mathbf{v}=(\lambda+\mu)\mathbf{v}$, if and only if $(A+\mu I)\mathbf{v}=(\lambda+\mu)\mathbf{v}$.
The fact that the determinant of $A+I$ is $0$ tells you that $0$ is an eigenvalue of $A+I$, and therefore, that $-1$ is an eigenvalue of $A$ (so that $-1+1=0$ will be an eigenvalue of $A+I$).
Once you know that $-1$ is an eigenvalue of $A$, you know that the other two eigenvalues add up to $1$ (so the trace will be $0$), and have a product of $-6$ (so the determinant will be $6$). So you are looking for $r_1$ and $r_2$ with $r_1+r_2=1$, $r_1r_2=-6$. This gives you the quadratic $(\lambda-r_1)(\lambda-r_2) = \lambda^2 -\lambda - 6$ whose roots are $3$ and $-2$, giving you the other two eigenvalues.