Factorizing, $p^m-n^3=8 \implies p^m=(n+2)(n^2-2n+4)$ Thus $n+2=p^a$ and $n^2-2n+4=p^b$ $\implies p^a(p^a-6)+12=p^b$ Now since, $a>0 (n+2>1)$ $p|12$ That is $p=2,3$ CASE 1 $p=2$ $2^a(2^{b-a}-2^a+6)=12$ Forming extensive cases of the factors on both sides (as $a=0,1,2$), we are done ! CASE 2 can be worked upon similarly.

utkarshgupta wrote: Factorizing, $p^m-n^3=8 \implies p^m=(n+2)(n^2-2n+4)$ Thus $n+2=p^a$ and $n^2-2n+4=p^b$ $\implies p^a(p^a-6)+12=p^b$ Now since, $a>0 (n+2>1)$ $p|12$ You showed $p|12$ from the third line.But eventually you didn't prove $a \ge b$.If $a\ge b$ then your conclusion is right.But if $a\ge b \implies p^a \ge p^b \implies n+2 \ge n^2-2n+4 \implies 3n \ge n^2+2$.It is true for only $n=1,2$.So your solution is partially right.

It's not that my solution is partial cause it was intended to be partial ! It's not a solution intended to be written in an exam. I have omitted such trivial cases ! Also my conclusion is right when $a \ge b$ and not when the opposite. Observe my solution carefully !

Nice diophantine, clearly $m$ has to be posititve integer. Case 1: $n=0$ $$p^m=8 \implies p=2 \; \text{and} \; m=3$$Case 2: $n$ is a posititve integer. Then clearly $n$ must be odd and also if we assume that $n \ne 1,2$ then by zsigmondy we have that there exists $q$ prime such that $q \not \; \mid n+2$ but $q \mid n^2-2n+4$ and using mod q on the diophantine we get $p=q$ and now using $p$-adic functions: $$m=v_p(n^3+8)=v_p(n^2-2n+4) \implies p^m \mid \mid n^2-2n+4 \implies n^3+8 \le n^2-2n+4 \implies n+2=1 \; \text{contradiction!!}$$Hence all the pairs are $(m,n,p)=(3,0,2), (2,1,3), (4,2,2)$ thus we are done

@ above p = 2 , m = 4 , n = 2 is also a solution.

I claim that $\boxed{(p, m, n) \in \{(2, 3, 0), (2, 4, 2), (3, 2, 1)\}}$ are the only solutions over $\mathbb{Z}^3_{\geq 0}$. Observe that we may rearrange and factor the initial equation as follows: \[p^m - n^3 = 8 \implies p^m = (n + 2)(n^2 - 2n + 4).\]Therefore we must have \[\begin{cases} n + 2 = p^a \\ n^2 - 2n + 4 = p^b, \text{ for integers a, b.} \end{cases}\]We may now split into two cases according to whether $a < b$ or $b < a$: Case 1: $(b < a)$. Then we must have $n + 2 > n^2 - 2n + 4 \implies n \leq 2$. Checking for $n = 0, 1, 2$ yields the solutions $(p, m) = (2, 3), (3, 2), (2, 4)$, respectively. $\square$ Case 2: $(b > a)$. Then we must have that \[\frac{n^2 - 2n + 4}{n + 2} = p^{b - a} \in \mathbb{Z} \implies (n - 4) + \frac{12}{n + 2} \implies n + 2 \mid 12 \implies n = 0, 1, 2, 4, 10, \]as the only valid values of $n$. As $n = 0, 1, 2$ have been undergone, we simply check $n = 4, 10$: \begin{align*} \underline{n = 4}: &\implies (4 - 4) + \frac{12}{4 + 2} = 2 \implies p = 2 \implies 2^m - 64 = 8 \implies \text{ no solution;} \\ \underline{n = 10}: &\implies (10 - 4) + \frac{12}{10 + 2} = 7 \implies p = 7 \implies 7^m - 1000 = 8 \implies \text{ no solution}. \square \end{align*} Therefore the only solutions are those solutions initially claimed. $\blacksquare$