Consider the multivariate polynomial $$f(x_1,\ldots,x_m)=mk\sum_{i=1}^mx_i^2mk(k1)\sum_{i=1}^mx_i\left(\sum_{i=1}^mx_i\right)^2,$$ for integers $m,k\ge2$. We are looking for integral zeros of $f$ with $0\le x_i\le k$. If $mk$ is square free, then it is easily seen that the only zeros of $f$ under the above conditions are $(0,\ldots,0)$ and $(k,\ldots,k)$. We believe that given $k$ if $m$ is large enough and $mk$ is divisible by square of an odd prime, then $f$ has more zeros. Could somebody provide a proof?

1$\begingroup$ Doesn't the sufficient magnitude of $m$ depend on the odd prime? $\endgroup$ – Max Alekseyev Nov 15 '19 at 20:52
First, I will consider the case of $k\geq 4$.
Let $mk=aq^2$, where $q$ is an odd prime. Since $f(x_1,\dots,x_m)=0$ implies that $mk$ divides $(\sum x_i)^2$, we will look for a zero with $\sum x_i = aq$. Then $f(x_1,\dots,x_m)=0$ will follow from the two equations: $$\begin{cases} \sum_{i=1}^m x_i = aq, \\ \sum_{i=1}^m x_i^2 = (k1)aq + a. \end{cases} $$ We will show that there is a solution containing only $0$'s, $1$'s, $2$'s, $(k1)$'s and $k$'s, i.e. $$\begin{cases} u + 2v + (k1)w + kt = aq,\qquad(\star) \\ u + 4v + (k1)^2w + k^2t = (k1)aq + a, \end{cases} $$ where $u,v,w,t$ are the multiplicities of $1$'s, $2$'s, $(k1)$'s, $k$'s, respectively.
This system has an integer solution whenever \begin{split} (k1) &\mid k^2(aqu2v)  k((k1)aq + a  u  4v),\\ k&\mid (k1)^2(aqu2v)  (k1)((k1)aq + a  u  4v), \end{split} that is $$\begin{cases} 2v \equiv  a(q1) \pmod{k1},\\ 2u + 6v \equiv a\pmod{k}. \end{cases}$$
We fix integers: \begin{split} \text{if $k$ is even:} & \qquad\begin{cases} v := \frac{a(q1)}2\bmod (k1), \\ u := \frac{a}{2}  3v\bmod{\frac{k}2}, \end{cases} \\ \\ \text{if $k$ is odd:} & \qquad \begin{cases} v := \frac{a(q1)}2\bmod \frac{k1}2, \\ u := \frac{a}{2}  3v\bmod{k}, \end{cases} \end{split} to further obtain integers: \begin{split} w &:= \frac{a(q1)(k1)u2(k2)v}{k1},\\ t &:= \frac{a+(k2)u + 2(k3)v}k. \end{split}
From definition of $u,v$, we have that they are smaller than $k$. It further follows that for a fixed $k$ and large enough $m$ (and thus large enough $aq$), values $w$ and $t$ are positive. From equation $(\star)$ it also follows that $u+v+w+t<m$. So, we constructed a zero of $f$ different from the trivial ones.
For $k=3$, the above construction works but $2$'s and $(k1)$'s collapse and have total multiplicity $v+w$. Similarly, for $k=2$, we get a solution with $1$'s and $2$'s having multiplicities $u+w$ and $v+t$, respectively.

$\begingroup$ Thanks Max for your nice argument. I just saw your answer as the internet was down this week in our country! $\endgroup$ – Ebrahim Nov 22 '19 at 14:33