Your test is in fact equivalent to the Lucas Lehmer test. First of all because of $$f(f(x))=T_4(x)$$ with $$f(x)=2x^2-1$$ your test is equivalent to the following test :
If $$S_0=2$$ and $$S_{i+1}=f(S_i)$$ then $$2^p-1$$ is prime if and only if $$S_{p-1}\equiv -1\mod M_p$$
If we compare to Lucas-Lehmer, we have $$T_0=4$$ $$T_{i+1}=T_i^2-2$$ then the statement is that $$2^p-1$$ is prime if and only if $$T_{p-2} \equiv 0\mod M_p$$ which is equivalent to $$T_{p-1}\equiv -2\mod M_p$$
We can easily show $T_i=2S_i$ for all $i$ by considering $$x^2-2=2(2(\frac{x}{2})^2-1)$$ and from this the equivalence easily follows.
As pointed out in the comments, your intuition should probably not be pointing at induction here, because the identities for $k$ and $k+1$ seem unconnected.
An arithmetic function is multiplicative if $f(ab)=f(a)f(b)$ for all coprime whole numbers $a$ and $b$ (if it is true for all whole numbers with no restrictions, it is completely multiplicative). It is instructive to show that if $f$ is multiplicative, then so too is the function $F(n):=\sum_{d\mid n}f(d)$. Indeed, more generally, if $f$ and $g$ are multiplicative then so too is their convolution $(f\ast g)(n):=\sum_{d\mid n}f(d)g(n/d)$. I really recommend grappling with this fact to see why it is true.
Both $\mu(n)^2$ and $J_k(n)$ are multiplicative, so $\mu^2/J_k$ is multiplicative, so $\sum_{d\mid n}\mu(d)^2/J_k(d)$ is too. As is the arithmetic function $n^k/J_k(n)$ on the other side of the equation.
Another important fact about arithmetic functions is they are determined by their values on prime powers. Thus, to prove $F(n)=G(n)$ for all $n$, it suffices to prove $F(p^v)=G(p^v)$ for all prime powers $p^v$. Often this simplifies things. For instance, in our case the identity becomes
$$ \frac{1}{1}+\frac{1}{p^k(1-\frac{1}{p^k})}+0+0+\cdots=\frac{p^v}{J_k(p^v)} $$
which is readily verifiable, so the claim is proved.
Best Answer
Suppose $2^p-1$ is composite, then it is not a prime power by Mihăilescu's theorem. Let $q,r\mid 2^p-1$ be two distinct prime factors, then both $q$ and $r$ are odd, whence $\varphi(q)$ and $\varphi(r)$ are both even. It follows that $$4\mid\varphi(q)\varphi(r)=\varphi(qr)\mid \varphi(2^p-1).$$