$2^mp^2=n^3-3^3=(n-3)\left(n^2+3n+9\right)$. $p\neq 2$, because $n^2+3n+9$ is always odd (and larger than $1$). If $p=3$, then $9\left(2^m+3\right)$ can't be a cube, because $2^m+3$ is not divisible by $3$. If $p\ge 5$, then $\gcd\left(n-3,n^2+3n+9\right)=1$, because $n^2+3n+9\equiv 3^2+3\cdot 3+9\equiv 27\pmod{n-3}$, so by Euclidean algorithm $\gcd\left(n-3,n^2+3n+9\right)=\gcd(n-3,27)\mid 27$ and $2^mp^2$ is not divisible by $3$. Also $n^2+3n+9$ is always odd, so $n^2+3n+9=p^2$ and $n-3=2^m$, so $(2n+3)^2+27=4p^2$, so $27=(2p+(2n+3))(2p-(2n+3))$ and $2p+(2n+3)>2p-(2n+3)+6$, so $2p+(2n+3)=27, 2p-(2n+3)=1$, so $(p,n)=(7,5)$, so $(p,m)=(7,1)$ is the only solution.

IstekOlympiadTeam wrote: Find all pairs $(m,p)$ of natural numbers , such that $p$ is a prime and \[2^mp^2+27\]is the third power of a natural numbers $2^mp^2=n^3-3^3=(n-3)(n^2+3n+9)=n(n+3)+9$. Note that, $n$ and $n+3$ are of different parity. So, one of them is even and $n^2+3n+9$ is odd. If $p=3$ then $p$ divides $n$, so $3^3$ divides $2^mp^2$, contradiction. Thus, $p\neq3$. If $p$ divides $n-3$ and $n^2-3n+9$, then $p$ divides $(n-3)^2-(n^2+3n+9)=9n$. But $p$ can not divide $3$ or $n$. Therefore, $p$ doesn't divides $n-3$. We get $n-3=2^m$ and $p^2=n^2-3n+9$. And now, we can finish it like the post above. gethd wrote: $n^2+3n+9=p^2$ and $n-3=2^m$, so $(2n+3)^2+27=4p^2$, so $27=(2p+(2n+3))(2p-(2n+3))$ and $2p+(2n+3)>2p-(2n+3)+6>6$, so no solutions.

$m = 1$ and $p = 7$ works. If $p$ is $2$, then we have $2^{m+2} = (x-3)(x^{2}+9+3x)$ Let $x-3 = 2^a$ and $x^{2}+9+3x = 2^{b}$ square the first equation and subtract from the second to get $9x = 2^{b} - 4^{a}$ Working modulo $2$ we see that if $a \geq 1$ and $b \geq 1$ then $x$ is even and we get a contradiction since $x = 3 + 2^{a}$. if $a = 0$ then $x = 4$ and $16 + 9 + 12$ is not a power of $2$. The other cases are: 1) $(x-3) = 2^{a}$ and $(x^2 + 9 + 3x) = 2^{b}p^{2}$ Substituting the value of $x$ from the first into the second, we get $4^{a} + 2^{a}*9 + 27 = 2^{b}p^{2}$ If $a \geq 1, b\geq 1$ then working modulo 2, we see that there are no solutions. $a = 0, b \geq 1$ gives us $37 = 2^{b}p^{2}$. Again no solutions. $b = 0, a=1$ gives us $p = 7$. hence one solution is $\left(m,p\right) = \left(1,7\right)$. $b = 0, a\geq1$ gives us no solution when we try modulo $4$. 2) $(x-3) = p*2^{a}$ and $(x^2 + 9 + 3x) = p*2^{b}$ substituting in second equation we get: $4^{a}p^{2} + 27 + 9p*(2^{a}) = p*2^{b}$ hence working modulo $p$ we see that $p = 3$ Plug $p = 3$ and then take modulo 3 to see that there are no solutions. 3) $(x-3) = p^{2}*2^{a}$ and $(x^2 + 9 + 3x) = 2^{b}$ Substitute the value of $x$ from the first into the second equation. $27 + 4^{a}p^{4} + 9*2^{a}*p^{2} = 2^{b}$ Working modulo $2$ we see that $a = 0$ or $b = 0$. But $b \geq 4$ in this case and so $a = 0$. Then working modulo $2$ tells us that there are no solutions. Hence the only solution i got is $\left(m,p\right) = \left(1,7\right)$.