It's a question I've been asking myself for a long time, without finding the answer. There are also variants, such as the sequence $2+6+8+10+12+14+\cdots$ which is in fact the same, but without the term $4$, as in the sequence $2+8+10+12+14+\cdots$
If anyone has the answer, I'm interested, thank you.
No, that sum is never a perfect square.
Arguing by contradiction, suppose there exist positive integers $m,n$ such that
$$2+4+...+2n = m^2.$$
Recall that
$$2+4+...+2n = 2(1+2+...+n)=2\left(\frac{n(n+1)}{2}\right)=n(n+1),$$
so the equality becomes
$$n(n+1) = m^2.$$
Note that $n$ and $n+1$ are consecutive integers, so in particular they're coprime. This forces both $n$ and $n+1$ to be perfect squares. That is, there exist positive integers $k,l$ such that $n=k^2$ and $n+1=l^2$. It follows that
$$k^2 +1 = l^2,$$
i.e. $$(l+k)(l-k) = l^2-k^2 = 1.$$
Since the only positive integer which divides 1 is also 1, we conclude that $l+k = l-k = 1$. Thus, $k=0$ and $l=1$, against both $k$ and $l$ being positive.