Can it be shown that there is no such pair exist for any $u,m>1$ such that
$$n^{2m}+(n-d)^{2m}+(n-2d)^{2m}+\cdots+(n-ud)^{2m}=a^{2m}$$
Where $n,u,d,m,a\in \mathbb{Z}_+$
There are lots of example easily found for $m=1$ but not getting for $m>1$.
Example for $m=1$, $1^2+(1-1\cdot2)^2+(1-2\cdot2)^2+(1-3\cdot2)^2=6^2$
Source code(pari\gp)
for(n=1,50,for(u=2,10,for(d=1,10,for(m=1,5,for(a=1,50,if(sum(q=0,u,(n-q*d)^(2*m))==a^(2*m),print([n,u,d,m,a])))))))
Edit
Related claim (which was already posted here)
Can it be shown that there is no such pair exist for any $m\ge 4$ such that
$$\sum_{q=0}^u(n+qd)^{m}=a^{m}$$
Where $n,u,d,m,a\in\mathbb{Z}_+$
By related claim allow, there is no such pair exist for $m>1$ such that
$$n^{2m}+(n+d)^{2m}+\cdots+(n+ud)^{2m}=a^{2m}$$