Numbers of the kind $0.aaa\ldots =\frac{1}{aaa\ldots a}$

I think a complete answer should cover all bases (no pun intended). We will have:

$$a = \frac {b^d-1}{\sqrt{b^{kd}-1}}$$

for any base $b$. However for $k\ge 2$:

$$\sqrt{b^{kd}-1}\ge \sqrt{b^{2d}-1} = \sqrt{(b^d-1)(b^d+1)} \ge b^d-1$$

forcing $a<1$. Hence we must have $k=1$, with $a = \sqrt{b^d-1}$, or $b^d - a^2 = 1$.

By Catalan's conjecture (or Mihăilescu's theorem, or if you are interested, this $1+a^2 = b^d$ is a special case proven by V. A. Lebesgue using Gaussian integers), if $d \ge 2$ there are no solutions. This forces $d = 1$, and we have only the uninteresting case:

$$1+a^2 = b$$

so our choice of $b$ must be one more than some square. For base $10$, $10 = 1+3^2$, so $0.\dot3 = \dfrac13$.