This is the solution for all integer numbers $m,n \ge 0$. The equation is equivalent to \[(m^2-n)(m^4+m^2n+n^2)=m-5n^2.\] Case 1. If $m=5n^2$ then $m^4+m^2n+n^2=0$ or $m^2-n=0$. If $m^4+m^2n+n^2=0$ then $m=n=0$. If $m^2=n$, we also have $m=5n^2$ then $25n^4=n$. We get $n=m=0$. Case 2. If $m>5n^2$. It follows that $m-5n^2 \ge m^4+m^2n+n^2$ or $m \ge m^4+m^2n+6n^2$. Since $m^4 \ge m$ so $m=n=0$, a contradiction. Case 3. If $m<5n^2$. With $m=0$ then $n^2(n-5)=0$. We obtain $n=5$. With $m \ge 1$, from the equation we get $\frac{5n^2-m}{m^4+m^2n+n^2}=n-m^2=k \ge 1$. The equation become \[\begin{array}{l} (5-k)n^2-nm^2k-m^4k-m=0 \qquad (1) \\ \Delta= (m^2k)^2+4(5-k)(m^4k+m)=m^4(20k-3k^2)+4m(5-k). \end{array}\] Equation $(1)$ has roots when $\Delta \ge 0$. It is easy to see that if $k \ge 7$ then $\Delta < 0$. Therefore $1 \le k \le 6$. If $k=6$ then $n=m^2+6$. Hence $6[m^4+m^2(m^2+6)+(m^2+6)^2]=5(m^2+6)^2-m$ or $m(13m^3+48m+1)=-36$, a contradiction since $LHS >0> RHS$. If $k=5$ then $n=m^2+5$. Hence $5[m^4+m^2(m^2+5)+(m^2+5)^2]=5(m^2+5)^2-m$ or $m(10m^3+25m+1)=0$, a contradiction since $LHS>0$. If $k=4$ then $n=m^2+4$. Hence $4[m^4+m^2(m^2+4)+(m^2+4)^2]=5(m^2+4)^2-m$ or $m(7m^3+8m+1)=2^4$. Since $\gcd (m,7m^3+8m+1)=1$ and $m<7m^3+8m+1$ so $2 \nmid m$. We find $m=1$. Hence $n=5$. If $k=3$ then $n=m^2+3$. It follows that $3[m^4+m^2(m^2+3)+(m^2+3)^2]=5(m^2+3)^2-m$ or $m(4m^3-3m+1)=18$. We see that $m \ne 1$. If $m \ge 2$ then $LHS>18$, a contradiction. If $k=2$ then $n=m^2+2$. It follows that $2[m^4+m^2(m^2+2)+(m^2+2)^2]=5(m^2+2)^2-m$ or $m(m^3-8m+1)=12$. Since if $m \ge 4$ then $LHS>RHS$ then $m \le 3$. We find $m=3$. Ir follows that $n=11$. If $k=1$ then $n=m^2+1$. Hence $m^4+m^2(m^2+1)+(m^2+1)^2=5(m^2+1)^2-m$ or $m(4m^3-7m+1)=4$. If $m \ge 2$ then $LHS>RHS$ and $m \ne 1$. Thus, the answer is $(m,n)=(0;0),(0;5),(1;5),(3;11)$. $\blacksquare$

Notice that for $n>11$ we have (with using $m<n$) $(3n-5)^3>(3m^2)^3=27(n^3-5n^2+m)>(3n-6)^3$ So no solution for $n>11$.

Another simple solution: from the condition, number m^6-m is of the form n^3-5n^2. For m>=4 we have that (m^2+1)^3-5(m^2+1)<m^6-m<(m^2+2)^3-5(m^2+2), so no solutions because n^3-5n^2 is strictly increasing in naturals which are >=5. After checking other cases we get that (m,n)={(1,5),(3,11)}.

shinichiman wrote: Thus, the answer is $(m,n)=(0;0),(0;5),(1;5),(3;11)$. $\blacksquare$ but $m, n \ne 0$