Here is what I observed :
Let $(2^p+1)/3$ be a Wagstaff number $W_n$.
Let the sequence $a_n = ord({a_{n-1}},\frac{2^{p}+1}{3}), a_0 = 2p$, where $p$ is a prime number.
Then $W_n$ is prime if the number $2$ is reached after some iterations.
For example :
With $W_{11}$ we get $2p = 22$, $(2^{11}+1)/3 = 683$
- $ord(22, 683) = 341$
- $ord(341, 683) = 11$
- $ord(11, 683) = 682$
- $ord(682, 683) = 2$
We got the sequence $22 \rightarrow 341 \rightarrow 11 \rightarrow 682 \rightarrow 2$, we reached $2$ so $683$ is prime.
Another example :
With $W_{29}$ we get $2p = 58$, $(2^{29}+1)/3 = 178956971$
We got the sequence $58 \rightarrow 1516584 \rightarrow 29 \rightarrow 1516584 \rightarrow 29$ etc., We don't reached $2$ so $178956971$ is not prime.
For the moment, I didn't find a counterexample.
I need help for proving it if it's true but I don't know how to start. If you found a counterexample please tell me.