Is there a Lucas-Lehmer equivalent test for primes of the form ${3^p-1 \over 2}$?

I refer you to Brillhart et al., Factorizations of $b^n\pm1$, published by the American Mathematical Society. The section, Introduction to the Main Tables, subsection Developments Contributing to the Present Tables, subsubsection Developments in Primality Testing, will tell you what methods there are for factoring such numbers.

There is the basic idea behind the Lucas-Lehmer primality test. For $n= 2^p-1$ we can choose $d= 3$ (quadratic reciprocity) and $\alpha= 2+\sqrt{3}$

If $n$ is prime and $d$ is not a square $\bmod n$ then $$\mathbb{Z}/n\mathbb{Z}[\sqrt{d}] = \{ a+b \sqrt{d}, (a,b) \in \mathbb{Z}/n\mathbb{Z}\}$$ is a field with $n^2$ elements, and its multiplicative group is cyclic with $n^2-1$ elements. If we find an element $\alpha$ of multiplicative order $n+1$ (ie. $\alpha^{n+1} \equiv 1 \bmod n, \alpha^{(n+1)/p} \not\equiv 1 \bmod n $ for every prime divisors $p | n+1$) then $n$ is prime.

(since otherwise with $q$ the least prime divisor of $n$ then $\mathbb{Z}/q\mathbb{Z}[\sqrt{d}]^\times$ is a group with less than $q^2-1 \le n-1$ elements, so the order of $\alpha\bmod q$ can't be $n+1$)

Thus all we need is to know the prime divisors of $n+1$ and compute $\alpha^{(n+1)/p} \bmod n$ for many $\alpha$. The same idea works in $\mathbb{Z}/n\mathbb{Z}$ if we know the prime divisors of $n-1$, and in $\mathbb{Z}/n\mathbb{Z}[x]/(f(x))$ for some irreducible polynomial $f$ of degree $k$ if we know a large part of the factorization of $n^k-1$.

$$\boxed{\ \ \text{Thus it doesn't work for }\ \frac{3^a-1}{2} \quad(\text{but it does for }2\cdot 3^a-1)\ \ }$$