You have $\overline{X}$ complete & sufficient and moreover $E[ \overline{X} ] = 1/\theta$; i.e. $\overline{X}$ is the UMVUE for $1/\theta$. It seems reasonable to guess that $1/\overline{X}$ may be the UMVUE for $\theta$. Note that $\sum_{i=1}^n X_i \sim \Gamma(n,\theta)$ since each $X_i$ is exponential rate $\theta$ and they're iid. Let $Z \sim \Gamma(n,\theta)$.
\begin{align*}
E[1/\overline{X}] = n E[1/Z] &= n \int_0^\infty \dfrac{1}{z} \dfrac{\theta^n}{\Gamma(n)} z^{n-1} e^{- \theta z} \; dz \\
&= n \int_0^\infty \dfrac{\theta^n}{\Gamma(n)} z^{n-2} e^{-\theta z } \; dz \\
&= n \theta \dfrac{\Gamma(n-1)}{\Gamma(n)} \underbrace{\int_0^\infty \dfrac{\theta^{n-1}}{\Gamma(n-1)} z^{n-2} e^{-\theta z } \; dz}_{=1} \\
&= \dfrac{n \theta \Gamma(n-1)}{\Gamma(n)} = \dfrac{n \theta}{n-1}
\end{align*}
So $ \dfrac{n-1}{n} \cdot \dfrac{1}{\overline{X}} = \dfrac{n-1}{\sum_{i=1}^n X_i}$ is the UMVUE for $\theta$.
Having that
$$\theta^m=P\{ X_1=x_1,X_2=x_2,...,X_m=x_m\}$$
An unbiased estimator for $\theta^m$ is
$$T=
\begin{cases}
1, & if \ \ X_1=X_2= \, ... \,=X_m =1 \\
0, & in \ other \ case
\end{cases}$$
But
$$\begin{align} E[T|S=s] & = P\{X_1=1,X_2=1,...,X_m=1|S=s\}=\frac{P\{X_1=1,X_2=1,...,X_m=1,S=s\}}{P\{S=s\}} = \\\\ & = \begin{cases}
0, & if \ \ m>s \\
\frac{\theta^m\binom{n-m}{s-m}\theta^{s-m}(1-\theta)^{n-s}}{\binom{n}{s}\theta^s(1-\theta)^{n-s}}, & if \ \ m\leq s
\end{cases} \end{align}$$
By the theorem of Lehmann-Scheffé, the UMVUE for $\theta^m$ is, after operating the latter expression:
$$E[T|S=s]=\begin{cases}
0, & if \ \ m>s \\
\frac{s!(n-m)!}{n!(s-m)!}, & if \ \ m\leq s
\end{cases}$$
Best Answer
Indeed, for a sample $(X_1,X_2,\ldots,X_n)$ from the $U(0,\theta)$ distribution, a complete sufficient statistic for the family is $$T(X_1,X_2,\ldots,X_n)=X_{(n)}$$
, the density of $T$ being $$g_{\theta}(t)=\frac{n}{\theta^n}t^{n-1}\mathbf1_{0<t<\theta}$$
So by Lehmann-Scheffe, an unbiased estimator of $e^\theta$ based on $T$ will be the UMVUE of $e^\theta$.
Let $h(T)$ be the required unbiased estimator.
Then, for all $\theta>0$,
\begin{align} \qquad\quad\frac{n}{\theta^n}\int_0^{\theta}h(t)t^{n-1}\,dt&=e^\theta \\\implies \int_0^{\theta}h(t)t^{n-1}\,dt &= \frac{\theta^ne^\theta}{n} \end{align}
Differentiating both sides wrt $\theta$,
\begin{align} h(\theta)\theta^{n-1}&=\frac{(n+\theta)\theta^{n-1}e^\theta}{n} \\\implies h(\theta)&=\frac{(n+\theta)e^{\theta}}{n} \end{align}
Hence your UMVUE must be $$h(T)=\frac{(n+T)e^T}{n}$$
This method to find the UMVUE works assuming that the UMVUE already exists.