Unique solution

$\begin{aligned} p^{2x+1} + 1 & = \sqrt{44} \color{#3D99F6}{e^{x\ln{p}}} \\ \color{#D61F06}{p^{2x+1}} + 1 & = \sqrt{44} \color{#3D99F6}{p^{x}} \\ \color{#D61F06}{p \dot{} p^{2x}} + 1 & = \sqrt{44} p^{x} \quad \quad \color{#3D99F6}{\text{Let } y = p^x} \\ py^2 - \sqrt{44} y + 1 & = 0 \\ \Rightarrow y & = \frac{\sqrt{44} \pm \sqrt{\color{#3D99F6}{44-4p}}}{2p} \end{aligned}$

For a unique $x$ and hence a unique $y$ , $\quad \Rightarrow \color{#3D99F6}{44 - 4p} = 0 \quad \Rightarrow p = \boxed{11}$ .