This is pretty straightforward; we just use the relationship between the Poisson and the chi squared:
If $Y\sim \text{Poisson}(\lambda)$ and $X\sim \chi^2_{2(k+1)}$, for integer $k$, then
$$F_Y(k) = 1-F_{X}(2\lambda) \,.$$
As a result, $$\lambda = \frac{1}{2}\, F_{X}^{-1}(1-F_Y(k))\,.$$
For example, in R, let's try to find the value of $\lambda$ corresponding to $k=6$ and $\alpha=0.1$:
> alpha=.1;k=6
> qchisq(1-alpha,2*(k+1))/2
[1] 10.53207
> ppois(k,10.53207)
[1] 0.1000001
So $\lambda\approx 10.53207$.
On my kids little laptop, running Rrelatively slow R*, $10^5$ such calculations took me a well under a second.
*(compared to C, say)
Hopefully that will be fast enough for you.