The equation HAS a solution - $(0, 0, 0)$. We will prove that it doesn't have any others. From the Fermat Little Theorem: $p|a^{2p+1}-a = a(a^p-1)(a^p+1)$ so for every integer $a$ we have $a^p \equiv 0$ or $1$ or $-1$ mod $p$. By verifying manually all cases we can see that the only possibility is $p=3$ (since $x^p+2y^p+5z^p \equiv 0 \mod{2p+1}$). So are left with proving that the equation $x^3+2y^3+5z^3 = 0$ has no solutions in integers. To do this consider mod $9$. One can easily check that we must have $x \equiv y \equiv z \equiv 0 \mod{3}$ but in this case just divide by $27$ and repeat the process. It will terminate unless $x=y=z=0$.

TomciO wrote: The equation HAS a solution - $(0, 0, 0)$. Good point. I have altered the wording.

let (x,y,z)=1,so multipling the equation and (x^p + 2y^p - 5z^2 ) and arguing on divisibility of x,y,z by 2p+1. the only case comes when p=3.(by using Fermat's little theorem). (x,y,z) = (1,1,2) or (2,1,1) module 3. now take mod 9 and you will get there's no answer.