Not a bad question! Mostly mathy in answer though.
Recharacterizing circles
So a circle is defined in the upper half-plane by the curve
$$
y(x) = \sqrt{R^2 - x^2}
$$
And since for small k, $\sqrt{1+k}\approx 1+\frac12 k,$ we have that for small $x$ a circle can be approximated by the parabola
$$
y(x) \approx R - \frac{x^2}{2R}.
$$
Now you can invert your perspective on all of this, you can say that a curve has a radius of curvature given by $[y''(x)]^{-1}$ and that circles are precisely those planar curves (they must be stuck in a 2D plane) which keep a constant radius of curvature (caveat: also not flip-flopping in whether it curves clockwise then anticlockwise in the plane, but you could say that the radius of curvature becomes undefined at such points).
Essentially, the radius of curvature tells us, if we want to approximate a curve with a little segment of circle, how big should that circle be.
Perpendicular acceleration
If we now graduate to $x(t), y(t)$ we have parametric motion in a plane, without loss of generality we can choose coordinate axes such that at some time $t_0$ the particle happens to be travelling in just the $x$-direction, $\dot x(t_0)=v, ~~\dot y(t_0) =0.$ If $\ddot x = a_\parallel$ and $\ddot y=a_\bot$ are nontrivial then we can expand the motion about $t=t_0$ as approximately,
$$
x(t) =x_0+ v~(t-t_0) + \frac12 a_\parallel (t-t_0)^2,\\
y(t) = y_0+ \frac12 a_\bot (t-t_0)^2.$$
One can do a quadratic formula for $t-t_0$ in terms of $x-x_0$ then substitute into the $y$-equation and take two derivatives to find, I believe,
$$
y''(x) = {a_\bot ~ v\over (v^2 + 2 a_\parallel(x-x_0))^{3/2}},
$$
so when we look for the radius of curvature at $x=x_0$ the tangential acceleration $a_\parallel$ is irrelevant and only the perpendicular component $a_\bot$ remains, moreover we find,
$$
R = \frac{v^3}{v~a_\bot} = \frac {v^2}{a_\bot}.$$
So this familiar formula for circular motion, if you are careful to apply it ONLY to the perpendicular acceleration component $a_\bot$, actually can be used in reverse to calculate a radius of curvature for a parametric path at some time $t_0$.
Assembling the argument
The argument is now the previous two sections put together:
If $\mathbf B$ is in the $z$-direction and $v_z=0$ is an initial condition, we never generate a $z$-component of acceleration and motion remains confined to the $xy$-plane.
There is never a tangential acceleration from a magnetic force, the cross product forces it to be perpendicular to velocity, so the velocity maintains an overall constant magnitude.
As a result $v^2$ is constant and $a_\bot$ is constant, so the radius of curvature is constant.
This finally implies that the motion is circular.
