A general conic is defined by five independent parameters and can pass through five arbitrary points.
Restricting to a parabola sets a constraint on the coefficients (the discriminant of the second degree terms must be zero), which "consumes" one degree of freedom.
But four remain, and you have an infinity of parabolas by the three given points and a fourth free one.
A more difficult question is when the shape of the parabola is fixed, i.e. you can only translate it and rotate it. Then it has only three degrees of freedom and the number of solutions must be finite. In the case of the vertices of an equilateral triangle, there can be at least six of them, by symmetry, as the figure shows.
In the general case, let the parabola have the equation $x=ay^2$, where $a$ is fixed. Then integrating the rigid transform, we need to solve the system
for $\theta, t_x$ and $t_y$.
By subtraction, we can eliminate $t_x$ and we get two equations linear in $t_y$.
Then eliminating $t_y$, we obtain a cubic polynomial equation in $\cos\theta$ and $\sin\theta$. We can rationalize it with the transform
This turns the trigonometric equation in a sextic one, having up to six real solutions.
The detailed discussion of the number of real roots seems to be an endeavor. As the minimum radius of curvature is $2a$, when the circumscribed circle of the triangle is smaller than this value, there is no solution.