Since $X \sim \operatorname{B}(1,k)$ is equal in law to $1-U^{1/k}$, where $U$ is uniform on a unit interval, you are computing $ \mathbb{P}\left( \prod_{i=1}^t (1-U_i^{1/k}) > x \right) $ The case of $k=1$ is special, due to $1-U \stackrel{d}{=} U$, and $-\log(\prod_{i=1}^t U_i) \sim \Gamma(t,1)$, as you mention.
Observe that $m_r = \mathbb{E}\left( B_t^r \right) = \mathbb{E}\left( B_1^r \right)^t = \binom{k+r}{r}^{-t}$. This permits to compute the moment generating function for $B_t$ in terms of the confluent generalized hypergeometric function: $ \mathcal{M}_{B_t}(u) = \mathbb{E}\left( \mathrm{e}^{u B_t} \right) = \sum_{r=0}^\infty \frac{u^r}{r!} m_r = {}_t F_t\left(\left. \begin{array}{c}\underbrace{1,\ldots,1}_{t\text{ times}} \\ \underbrace{k+1,\ldots,k+1}_{t\text{ times}} \end{array}\right| u \right) $ So in principle one can construct Chernoff bound. But it is known that moment bounds are tighter. Assuming $0 < x<1$, $ \mathbb{P}(B_t \geqslant x) \leqslant \inf_{r\geqslant 0} \left( \frac{m_r}{x^r}\right) = \inf_{r\geqslant 0} \left( \frac{1}{x^r} \binom{k+r}{r}^{-t}\right) = \left. \frac{1}{x^r} \binom{k+r}{r}^{-t} \right|_{r = \lfloor k \left(\frac{1}{1-x^{\frac{1}{t}}}-1\right)-1 \rfloor} $
