There are some variations of the question. Fixing the finite field ${\mathbb{F}}_q$, the number of all monic irreducible polynomials $x^2 - ax + b \in {\mathbb{F}}_q[x]$ is $(q^2 - q)/2$, which is easy to see just by dividing the number of elements in ${\mathbb{F}}_{q^2} \setminus {\mathbb{F}}_q$ by $2$.
The assumption $b = 1$ forces ${\mathrm{N}}(\alpha) = 1$ for a root $\alpha \in {\mathbb{F}}_{q^2} \setminus {\mathbb{F}}_q$ of the polynomial. Beyond this point my arguments about this turns out to be circular around and leading nowhere.
I tried using the fact: the radical $a^2 - 4$ is not a square in ${\mathbb{F}}_q$ and was hoping that the preimage of $4$ by the surjective map $\varphi : {\mathbb{F}}_q \times {\mathbb{F}}_q \rightarrow {\mathbb{F}}_q$ given by $(a,b) \mapsto -a^2 + b^2$ give something.
The question came up while counting the centraliser of an anisotropic regular semi-simple class in ${\mathrm{SL}}(2,{\mathbb{F}}_q)$.
Any help?
The number of reducible polynomials of the form $x^2+ax+1$ is easier to compute. This is the number of polynomials $(x-b)(x-b^{-1})$. For each $b$ you obtain a polynomial, but you count each one twice ($b$ and $b^{-1}$). However, if $b=b^{-1}$ you don't need to double-count. For $q$ odd there are two values for which is is true ($\pm 1$) and for $q$ even there is one ($1$).
Thus for $q$ odd you obtain $(q-3)/2+2$ reducible polynomials and for $q$ even you obtain $(q-2)/2+1$ reducible polynomials. Subtract this from $q$ to obtain the number of irreducible ones.
Thank you to Jyrki Lahtonen for pointing out that I'd been very hasty in my formulae earlier.