Let $f(k)=k^2+k+1$. Note the identity $k^4+k^2+1=(k^2-k+1)(k^2+k+1) = f(k)f(k-1)$, the expression reduces to
$$f(0)f(1)^2f(2)^2...f(n-1)^2f(n)$$thus it remains to show that $f(0)f(n) = f(n)$ is not a perfect square. But this is obvious as $n^2 < n^2+n+1 < (n+1)^2$.
Notice that $(1^4+1^2+1)(2^4+2^2+1)\dots(n^4+n^2+1)=\prod_{i=1}^{n}(i^4+i^2+1)=\prod_{i=1}^{n}(i^2-i+1)(i^2+i+1)=\prod_{i=1}^{n}(i^2-i+1)((i+1)^2-(i+1)+1)=\Big[\prod_{i=2}^{n-2}(i^2-i+1) \Big]^2(1^2-1^1+1)(n^2+n+1)$
Now it's suffices to show that $(1^2-1^1+1)(n^2+n+1)=n^2+n+1$ is not a perfect square for all positive integers $n$. Indeed, $n^2<n^2+n+1<(n+1)^2$. $\square$