Say that there is one such $a_{0}$ for which the sequence consists entirely of primes. The given recurrence can be rewritten as
$$ a_{n+1}+1 = 2(a_{n}+1) $$
Say that $a_{n}+1=b_{n}$
$$ \implies b_{n+1} = 2 b_{n} $$
So, the sequence $ \lbrace b_{i}\rbrace_{i \ge 0} $ is a Geometric progression. And therefore,
$$ b_{n} = b_{0} 2^{n} = (a_{0}+1)2^n = a_{n}+1 \implies a_{n} = (a_{0}+1)2^n -1 $$
$$ \implies a_{n} = 2^n a_{0} + 2^n-1 $$
Since $a_{0}$ is also a prime, so let $a_{0}=p$(a prime number).
$$ a_{p-1} = 2^{p-1} p + 2^{p-1}-1 $$
Using Fermat's $2^{p-1}-1$ is divisible by $p$ and $ 2^{p-1} p$ is divisible by $p$, therefore $a_{p-1}$ is divisible by $p$ and hence cannot be a prime. So, this sequence cannot output only primes.
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