$p \geq 5$ prime. $2p+1$ not a prime. Then $\phi(n)=2p$ has no solution.

In the case that $a=0,1$, you have used that $2p+1$ is not prime when you said that $\phi(n) = p_1^{\alpha_1 -1}(p_1 -1) = 2p$ gives $p_1-1 = 2$ and $p = p_1^{\alpha_1 - 1}$. Indeed we could instead have $p_1-1 = 2p$, and $\alpha_1 = 1$. But this is ruled out since then $p_1 = 2p+1$ is not prime.

You should also note that we could, a priori, have $p_1 - 1=p$, but then $p_1$ is not prime since $p\neq 2$.