nice problem/hint[hide]Hint: Factorize and divide in cases of x=0, x=even, x=odd. When x=even number prove that $p=5$.[/hide] Answer

Factor the equation to get $(x+3)(2x+1)=p^n$ If $x=0$, we can easily derive that $p=3,n=1$. If $x > 0$, Let $x+3=p^a$ and $2x+1=p^b$ for $a,b \in\mathbb{N}$ Clearly $a \le b$. If $a=b \implies p^a=p^b \implies x+3=2x+1 \implies x=2$ $\implies p=5,n=2$ If $a<b$, $p^b-2p^a=-5 < 0$ $\implies p^b < 2p^a$ $\implies p^a < p^b < 2p^a$ $\implies 1 < p^{b-a} < 2$. As no natural number lies between 1 and 2(both exclusive), a contradiction! Hence $(x,p,n) \equiv (0,3,1),(2,5,2)$ are the only solutions.