Edit: the first part of this isn't correct, see the comments from Zach Hunter below. I misinterpreted the question.

We can partition the integers into squares and non squares, so it would be sufficient to show that the squares can be partitioned with no sums. See this MathOverflow post from a little while ago for a discussion on partitioning the integers without Pythagorean triples.

It seems that this is an open problem, and only in 2016 was it shown that the squares cannot be 2-coloured to avoid monochromatic Pythagorean triples. If the integers were partitioned to avoid square sums, then certainly the squares would be, and the existence of a finite partition is an open problem. So, I believe this problem is open!

**Previous answer:**

This is unfortunately not a complete answer, but a collection of ideas.

A sufficient condition would be a colouring such that there are no monochromatic solutions to $x+y=z$. Unfortunately, this does not exist: by Schur's theorem, any partition of $\mathbb{Z}^+$ into finitely many parts, one part contains $x,y,z$ such that $x+y=z$. One way to try and proceed with this could be to try and modify a proof of Schur's theorem.

Recently, Bloom and Maynard showed the following density result: if $A \subseteq [N]$ has no solution to $a-b=n^2$ where $a,b \in A$ and $n \geq 1$, then $|A| = O\left( \frac{N}{(\log N)^{c \log \log \log N}}\right)$. Furthermore, $|A+A|^{3/4} \leq |A-A| \leq |A+A|^{4/3}$ is known, suggesting that the situation for $A+A$ to be square free should not be 'that different' from $A-A$ being square free.

My guess is that the answer to your question is no, but I cannot prove it. Perhaps someone with more knowledge than me will be able to answer properly, but until then I hope these comments are somewhat helpful for investigation.