Prove that the prime $p$ can only be $13$

You have

$$\begin{equation}\begin{aligned} 4p_1 + 1 & = 2p_2 - 1 \\ 4p_1 & = 2p_2 - 2 \\ 2p_1 & = p_2 - 1 \\ p_2 & = 2p_1 + 1 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

With $p_1$, consider its possible values modulo $3$. If it's $p_1 \equiv 1 \pmod{3}$, then $p_2 \equiv 0 \pmod{3}$, which is not allowed since $p_2 \gt 3$. Alternately, if $p_1 \equiv 2 \pmod{3}$, then $p_2 \equiv 2 \pmod{3}$ so $p = 2p_2 - 1 \implies p \equiv 0 \pmod{3}$. The only case where this may be possible is where $p_2 = 2$ giving $p = 3$, but then $p = 4p_1 + 1$ can't hold. This leaves the only possible case where $p_1$, $p_2$ and $p$ are all prime is where $p_1 = 3$, leading to your one case where $p = 13$.

Suppose $p\equiv1\bmod3$, then it is easy to verify that $\frac{p-1}4\equiv0\bmod3$, so $\frac{p-1}4=3$ and $p=13$.

Suppose $p\equiv2\bmod3$, then by similar logic $\frac{p+1}2\equiv0\bmod3$ and $p=7$, but then $\frac{p-1}4$ is non-integral.

Since $p>3$ by $\frac{p-1}4$ being prime, $p=13$.