Let $t=T_F$.  Keeping in mind that $t_1,t_2,\dots,t_N$ are unif$(0,t)$ given the number of occurrances $N=n$, the likelihood becomes
\begin{align}
L(\lambda,t) &= P(N=n) f(t_1,t_2,\dots,t_N|N=n) \\
&= \frac{e^{-\lambda t}(\lambda t)^n}{n!}(\frac1t)^n \\
&\propto e^{-\lambda t}\lambda^n.
\end{align}
for $t\ge t_\text{max}$ and zero elsewhere.  This is maximised for $\hat t=t_\text{max}=\operatorname{max}(t_1,t_2,\dots,t_n)$ and $\hat\lambda=n/t_\text{max}$.  These MLEs don't exist if there are no occurrences $N=0$, however.  Conditional on $N=n$, using the fact that the $t_i$'s are uniform, $E(t_\text{max}|N=n)=\frac n{n+1} t$.  Hence, the estimator $t^*=\frac {n+1}n t_\text{max}$ is unbiased conditional on $N=n$ and hence also conditional on $N\ge 1$.   A reasonable frequentist estimator of $\hat\lambda$ might be $\lambda^* = n/t^* = \frac{n^2}{(n+1)t_\text{max}}$ but this does not have finite expectation when $N=1$ so assessing its bias is even more troublesome.  

Bayesian inference using independent, non-informative scale priors on $\lambda$ and $t$ on the other hand leads to a posterior 
$$
f(\lambda,t|t_1,\dots,t_N) \propto e^{-\lambda t}\lambda^{n-1}t^{-1}.
$$
for $t>t_{max},\lambda>0$.  Integrating out $\lambda$, the marginal distribution of $t$ becomes
$$
f(t|t_1,\dots,t_N) = \frac{n t_\text{max}^n}{t^{n+1}}, t>t_\text{max}.
$$
with posterior mean $E(t|t_1,\dots,t_N)=\frac n{n-1} t_\text{max}$.  A $(1-\alpha)$-credible interval for $t$ is given by $\left(\frac{t_\text{max}}{(1-\alpha/2)^{1/n}}, \frac{t_\text{max}}{(\alpha/2)^{1/n}}\right)$.

The marginal posterior of $\lambda$,
\begin{align}
f(\lambda|t_1,\dots,t_N) &\propto \int_{t_\text{max}}^\infty e^{-\lambda t}\lambda^{n-1}t^{-1} dt \\
&= \lambda^{n-1}\Gamma(0,\lambda t_\text{max})
\end{align}
where $\Gamma$ is the incomplete gamma function.