I received an answer $(2;2)$, and proved that he is the only, but this solution is not short

If $n=2k+1$ then $n^5+n^4=M_8 +2$ but $7^m-1$ can be $M_8$ or $M_8 +6$. So we need $n=2k$. Then $m$ must be even($m=2j$). $16*k^4*(2k+1)=(7^j-1)(7^j+1)$. Since $gcd(k^4;2k+1)=1 , gcd(7^j-1;7^j+1)=2$ we get 4 cases. I'm gonna solve only one, cause the others can be solved similarily: $7^j-1=2*k^4*a ; 7^j+1=8*(2k+1)b$ , where $ab=2k+1$ and $gcd(a,b)=1$. Then $8*(2k+1)*b=2*k^4*a+2$ so if we multiply this by $a$ and simplify by 2 we get:$*4(2k+1)^2=2*k^4*a+2$. If $a\geq 10$ then$RHS>LHS$. We analyze the small cases and we'll get that the only solutions are $(0,0);(2,2)$

GGPiku wrote: If $n=2k+1$ then $n^5+n^4=M_8 +2$ but $7^m-1$ can be $M_8$ or $M_8 +6$. So we need $n=2k$. Then $m$ must be even($m=2j$). $16*k^4*(2k+1)=(7^j-1)(7^j+1)$. Since $gcd(k^4;2k+1)=1 , gcd(7^j-1;7^j+1)=2$ we get 4 cases. I'm gonna solve only one, cause the others can be solved similarily: $7^j-1=2*k^4*a ; 7^j+1=8*(2k+1)b$ , where $ab=2k+1$ and $gcd(a,b)=1$. Then $8*(2k+1)*b=2*k^4*a+2$ so if we multiply this by $a$ and simplify by 2 we get:$*4(2k+1)^2=2*k^4*a+2$. If $a\geq 10$ then$RHS>LHS$. We analyze the small cases and we'll get that the only solutions are $(0,0);(2,2)$ $n=3$ is Counterexample for the fact $n^5+n^4\equiv 2 mod 8$ $3^5+3^4\equiv 4 mod 8$ I think solution must be like that $n^5+n^4+1$ can be factorizing. İ think the rest is not so difficult.

Any idea? EDITED:Guys I got a solution.If someone wants i can write.

factorising and gcd