If $M \in \mathcal{M}_2(\mathbb{R})$, we have
$$\det (M - xI_2) = x^2 - \operatorname{tr}(M)x + \det(M), \quad \forall x \in \mathbb{R}. \quad (1)$$
Let $A \in \mathcal{M}_2(\mathbb{R})$ be a matrix satisfying the hypothesis. From (1) we obtain
$$\operatorname{tr}(A^{2014}) = \operatorname{tr}(A^{2016}) = 0. \qquad (2)$$
Using the Cayley-Hamilton theorem, we obtain
$$A^2 - \operatorname{tr}(A)A + \det(A)I_2 = O_2. \tag{3}$$
We analyze several cases.

If $\operatorname{tr}(A) = 0$, then (3) implies $A^2 = -\det(A)I_2$, whence $A^{2014} = (-\det(A))^{1007}I_2$. From (2) it results $\det(A) = 0$.

If $\det(A) = 0$, then this time (3) implies $A^2 = \operatorname{tr}(A)A$. Inductively, $A^{n+1} = \operatorname{tr}^n(A)A$, $\forall n \in \mathbb{N}^*$. It follows that $\operatorname{tr}(A^n) = \operatorname{tr}^n(A)$, $\forall n \in \mathbb{N}^*$. As a special case, using (2), $0 = \operatorname{tr}(A^{2014}) = \operatorname{tr}^{2014}(A)$, whence $\operatorname{tr}(A) = 0$.

If $\operatorname{tr}(A) \neq 0$ and $\det(A) \neq 0$. From (3), $A^{2016} - \operatorname{tr}(A)A^{2015} + \det(A)A^{2014} = O_2$. Then $\operatorname{tr}(A^{2016}) - \operatorname{tr}(A)\operatorname{tr}(A^{2015}) + \det(A)\operatorname{tr}(A^{2014}) = 0$. Using (2) and our assumption, we get $\operatorname{tr}(A^{2015}) = 0$. From (3), it follows that
$$\operatorname{tr}(A^n) = \frac{1}{\operatorname{det}(A)} \left[ \operatorname{tr}(A) \operatorname{tr}(A^{n+1}) - \operatorname{tr}(A^{n+2}) \right], \quad n \in \mathbb{N}^*.$$
Thus, starting with $\operatorname{tr}(A^{2014}) = \operatorname{tr}(A^{2015}) = 0$, we obtain recursively $\operatorname{tr}(A^{2013}) = 0$, $\operatorname{tr}(A^{2012}) = 0$, ..., $\operatorname{tr}(A) = 0$, a contradiction. It follows that $\operatorname{tr}(A) = 0$ and $\det(A) = 0$.

Finally, from (3) it follows that $A^2 = O_2$, hence $A^n = O_2$, $\forall n \ge 2$. Then, for $n \ge 2$, we have $\det(A^n - I_2) = \det(-I_2) = 1 = \det(I_2) = \det(A^n + I_2)$. For $n = 1$, from (1), $\det(A - I_2) = 1 = \det(A + I_2)$.