As @DavidESpeyer suggests, I think you meant ${}+ a$ in place of ${}- a$. As @KevinCasto says, you are then looking for the order of $\begin{bmatrix} 1 & a \\ 1 & 1 \end{bmatrix}$ as an element of $\operatorname{PGL}_2(\mathbb F_p)$, i.e. (since its eigenvalues are $1 \pm \sqrt a$), the order of $1 \pm \sqrt a$ as an element of $\mathbb F_{p^2}^\times/\mathbb F_p^\times$. Certainly this order divides $p + 1 = \lvert\mathbb F_{p^2}^\times/\mathbb F_p^\times\rvert$; but, as @DavidESpeyer also suggests, it need not equal $p + 1$. Indeed, if $p = 5$ and $a = 2$, then the element has order $3$, since $\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}^3 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. (If you really did mean ${}- a$ instead of ${}+ a$, then the same example works; just pretend I took $a = 3$ instead of $a = 2$.)
EDIT: There was some discussion about what would happen if you had ${}- a$ in place of ${}+ a$. If $-a$ as well as $a$ is not a quadratic residue (i.e., if $-1$ is a quadratic residue), then the reasoning above shows that the order divides (but need not equal) $p + 1$. If $-a$ is a quadratic residue, then you are now looking for the order of $(1 + \sqrt{-a}, 1 - \sqrt{-a})$ as an element of the quotient of $\mathbb F_p^\times \times \mathbb F_p^\times$ by the diagonal copy of $\mathbb F_p^\times$. The order thus now divides $p - 1$. Taking $p = 7$ and $a = 3$, and observing that $\begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix}^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, shows that the order may be a proper divisor of $p - 1$.
