I have checked numerically and it seems true that for any three non-zero real matrices $A,B,C$ of $\dim = 4$, and a scalar $z$, it holds that
$$\det(z^2 A + z B + C) = \det(z^2 A - z B + C).$$
Is there a way to prove this in general or disprove with a counter example?
If $A=B=C=I_4$, one would get $(z^2+z+1)^4=(z^2-z+1)^4$ for all $z\in\mathbb{R}$, which is clearly false for all but finitely many real numbers. (In fact, this is false for all $z\in \mathbb{R}^\times$). A concrete counterexample is obtained for $z=1$.
Edit Following OP's comment, here is another series of counterexamples.
Fix an invertible matrix $P$, and set $A=P$, $B=zP$, $C=z^2 P$.
We get $\det(z^2A+zB+C)=(3z^2)^4\det(P)$, while $\det(z^2A-zB+C)=(z^2)^4\det(P)$.
These quantities are different for all $z\in\mathbb{R}^\times$ and all invertible $P$.
You can also find a different family of examples as follows.
Set $A=I_4$, $B=M$ (to be chosen) and $C=0$. Then the equality becomes $z^4\det(z I_4+ M)=z^4\det(zI_4-M)$. Then pick any non zero $z$ and $M$ such that $z$ is an eigenvalue of $M$ but $-z$ is not an eigenvalue of $M$. Then LHS will be nonzero while RHS will be zero.