If your entries are complex, take a diagonal matrix with all entries equal to the fourth roots of unity.
In general, I guess you want to have $A^4 -I =0$ , but not the factorization:
$(A^2+I)(A^2 -I )=0$ with either factor being zero.
So we could, in theory, have zero-divisors here, i.e., we can have the
product be zero without either factor being zero. But this is not
possible with entries over $\mathbb R$ , since $\mathbb R$ is a field.
Since you want to avoid $A^2=I$ , your
only option is having matrices $A$ with $A^2+I=0$ , which greatly simplifies
EDIT : there I something wrong in my answer that I am trying to understand.
Specifically, there are matrices $A$ with $A^4=I$ , but that satisfy neither
$A^2=I$ , nor $A^2 =-I$ . Specifically, the diagonal matrix $D(-1,-1,1)$.
I suspect that there are factorizations for $A^4-I$ other than $(A+I)(A-I)$