Following the carifying comments above, let's go with the second definition. The idea there is that the curve's a circular helix if there's an axis ($P + td$) such that the projection of the curve onto the plane perpendicular to the axis traverses a circle of radius $A$ with speed $AB$ in that plane. That seems to capture circular helix-ness.
Now suppose you have a curve with constant curvature and torsion. To prove it matches the definition, you need to come up with the axis and the constants $A$ and $B$. Well, if you think of a model circular helix around the $z$-axis, say, the normal vector always lies in the $xy$-plane. So you could "discover" the axis by taking any two nearby normals and computing their cross-product. Of course, the more nearby they are, the shorter the cross-product vector will be, so to get a nice unit vector for defining the axis, I'm implicitly suggesting that you say this:
Let $$\vec{d} = \lim_{h \to 0} \frac{1}{h} N(h) \times N(0)$$.
You can rewrite that as
\begin{align}
\vec{d} &= \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \times N(0) \\
&=
\left( \lim_{h \to 0} \frac{1}{h} (N(h) - N(0)) \right)\times N(0) \\
&= N'(0) \times N(0)
\end{align}
Now you can use the Frenet-Serret formulas to work out an explicit value for the vector $\vec{d}$. (Note: The result won't generally be a unit vector, but it'll have nonzero length, and then you can make it a unit vector; without the division by $h$, you'd end up with a zero vector.)
Going back to the model of a circular helix around the $z$-axis, you should be able to infer the radius of the circle from the constant curvature and torsion. (Should be 1/curvature, I believe, but I'd have to check whether torsion somehow sneaks in there). That lets you pick a point $P$ at which to start the axis:
$$
P = \gamma(0) + r N(0)
$$
where $\gamma$ is your curve, and $r$ is the radius you computed. For vectors $\vec{e}$ and $\vec{f}$, I'd recommend $N(0)$ and $\vec{d} \times N(0)$. Now all you have to do is show that everything matches up as needed.
Since you didn't ask for a complete proof, but instead for a definition of circular helix that would allow you to prove that constant-curvature-and-torsion curves were indeed circular helices, I think I've given you what you asked for. I can fill in details if necessary, but that might take away the fun from you.
Following comments and discussion
Here's a more complete answer that depends only on the curve $\alpha$, and its Frenet frame. Since comments reveal that you're teaching (or taught) a DG course, I'll let you do the computations here, and just post the main results.
- The helical curve
$$
H(s) = (r \cos us, r \sin us, v)
$$
has tangent vector
$$
T(s) = (-ru \sin us, ru \cos us, v)
$$
which is a unit vector only if
$$
r^2 u^2 + v^2 = 1.
$$
So I'll consider only triples $(r, u, v)$ that satisfy that condition. I'll also restrict to $r > 0$.
The derivative of $T$ is
$$
T'(s) = (-ru^2 \cos us, -ru^2 \sin us, 0)
$$
whose length is $ru^2$; by the Frenet formulas, this is the curvature, $\kappa$, of the helix, and the normal vector to the helix is
$$
N(s) = (-\cos us , \sin us, 0).
$$
The derivative of the normal is
$$
N'(s) = u(\sin us, -\cos us, 0).
$$
The binormal is
$$
B(s) = T \times B = (v \sin ut, -v \cos ut, ru).
$$
The torsion, by Frenet, is $\tau = N' \cdot B = uv$.
So for such helical curves, we have
\begin{align}
ru^2 &= \kappa \\
r^2u^2 + v^2 &= 1 \\
uv &= \tau.
\end{align}
We can solve these for $r, u, v$ to get
\begin{align}
u &= \sqrt{\kappa^2 + \tau^2}\\
v &= \frac{\tau}{\sqrt{\kappa^2 + \tau^2}} \\
r &= \frac{\kappa}{\kappa^2 + \tau^2}
\end{align}
Thus for any positive $\kappa$ and any $\tau$, we can find a unit-speed helical curve $s \mapsto H_{\kappa, \tau}(s)$ by using the values of $u, v, r$ above in $H$.
Now let's look at the constant-speed, constant torsion curve $\alpha$. Let $Q = \alpha(0)$, and let $u, v, r$ be derived from the known curvature and torsion as above.
$$
\newcommand{bT}{\mathbf T}
\newcommand{bN}{\mathbf N}
\newcommand{bB}{\mathbf B}
\newcommand{Tb}{ T_\beta}
\newcommand{Nb}{ N_\beta}
\newcommand{Bb}{ B_\beta}
\newcommand{bD}{\mathbf D}
\newcommand{bE}{\mathbf E}
$$
Let $\bT, \bN, \bB$ denote the unit tangent, normal, and binormal of $\alpha$ at $s = 0$.
Define
\begin{align}
\beta(s) &= (Q + r\bN) - r\cos(us)\bN + r\sin(us) \bE + vs \bD, \text{ where}\\
\bD &= \frac{1}{\|\bN \times N'(0)\|} \bN \times N'(0)\\
&= \frac{1}{\|\bN \times (-\kappa\bT + \tau \bB)\|} \bN \times (-\kappa\bT + \tau \bB)\\
&= \frac{1}{\|\kappa\bB + \tau \bT)\|} (\kappa\bB + \tau \bT)\\
&= \frac{1}{\sqrt{\kappa^2+ \tau^2}} (\kappa\bB + \tau \bT), \text{ and}\\
\bE = \bN \times \bD,
\end{align}
which you can work out in terms of $\bT, \bN,$ and $\bB$ similarly.
Then direct computation shows that
$\beta$ and $\alpha$ start at the same point.
$\beta$ has constant curvature $\kappa$.
3, $\beta$ has constant torsion $\tau$.
By the fundamental theorem of curves, $\beta$ must equal $\alpha$. Bur $\beta$ is evidently a helix with centerline $\bD$. (Indeed, it's the the result of rotating the "standard helix" $H(\kappa, \tau)$ by the matrix whose columns are $\bT, \bN, \bB$, and then translating by $(Q + r\bN)$.)
Best Answer
Of course the uniqueness theorem (Fundamental Theorem of Curves) will do it, since you know (or can easily show) that circular helices have constant curvature and torsion.
The intent of my hint was to then deduce that $\vec N(s)=\cos(ks)\vec c_1+\sin(cs)\vec c_2$ with $k^2=\kappa^2+\tau^2$ and $\vec c_1,\vec c_2$ fixed vectors. Show then that $\vec c_1$ and $\vec c_2$ must be orthogonal and show that you get circular motion in the plane they span. Can you finish from there?