What can be said about the order of short Weierstrass curves with $a=1$ and $b=0$?
The curve $E$ given by the equation $y^2 = x^3 + x$ is a Montgomery curve. Those are
are the form $By^2 = x^3 + Ax^2 + x$, so in this particular case we have $A=0$ and $B=1$.
The points of order $2$ have their $x$-coordinate as a root of $x^3 + x$. If $-1$ is a square over $\mathbf F_p$, then the points of order $2$ are $(0,0)$, $(\sqrt{-1},0)$ and $(-\sqrt{-1},0)$. Otherwise there is only $(0,0)$.
On those curves, the closest to a prime for the cardinality is $4\cdot\text{prime}$ for both the curve and its quadratic twist (such as the Goldilock curve), or $8\cdot\text{prime}$ and $4\cdot\text{prime}$ (such as Curve25519).