Let $c_n$ denote the colour of positive integers $x$ such that $2^n \le x < 2^{n+1}$. We will determine colours $c_n$ inductively. First, let us choose $c_0$, $c_1$ and $c_2$ to be three different colours. Next, for each $n \ge 3$ let $c_n$ be the colour different from $c_{\lfloor n/2 \rfloor}$ and $c_{\lfloor n/2 \rfloor+1}$. Note that this is well-defined since $\lfloor n/2 \rfloor + 1 < n$ for all $n \ge 3$.
By construction condition i) holds. Let us prove that condition ii) also holds. Let $x$, $y$ and $z$ be positive integers of the same colour such that $x + y = z^2$. Without loss of generality we may assume that $x \le y$. Let $n$ be an integer such that $2^n \le y < 2^{n+1}$. Then we obviously have
$$2^n < x + y < 2^{n+2},$$
and hence $2^{n/2} < z < 2^{(n+2)/2}$. It follows that $2^{\lfloor n/2 \rfloor} < z < 2^{\lfloor n/2 \rfloor+2}$. Since $y$ and $z$ have the same colour, by construction it follows that $n \le 2$. From here we conclude $x \le y < 8$ and $z = 2$ or $3$. Direct verification shows that the only possibility is $x = y = z = 2$.