Look at $\tilde{T}(n) = T(2^n)$. The recurrence for $\tilde{T}$ is then $\begin{eqnarray} \tilde{T}(n) &=& \tilde{T}(n-1) + 2^n \\ T(0) &=& 1. \end{eqnarray}$
It follows that $\tilde{T}(n) = 1 + 2 + 4 + \ldots + 2^n = 2^{n+1} - 1$ and thus that $T(n) = 2^{\log_2(n)+1} - 1 = 2n - 1$ (if $n$ is a power of two, your recurrence doesn't define $T(n)$ for other values of n). A possible $\theta$-bound for $T$ is thus $n$, i.e. $T(n) \in \theta(n)$. Note that this assumes that $T(n)$ where $n$ isn't a power of two (a case which your recurrence doesn't define!) obeys the bound (derived soley from the power-of-two values) also. A natural assumption which guarantees that is $T(n) \leq T(n+1)$.