For $n = 12$ there exists a coloring of the numbers from $\{2, 3, \dots, 12\}$ with two colors such that the equation $x + y = z$ has no monochrome solution: we color the numbers from $A = \{2, 3, 4, 11, 12\}$ with one color, and the elements of $B = \{5, 6, 7, 8, 9, 10\}$ with the second color.
For $n < 12$ we color with the first color the numbers from $\{2, 3, \dots, n\} \cap A$ and with the second one the elements of $\{2, 3, \dots, n\} \cap B$. Again, the equation $x + y = z$ has no monochrome solution.
It follows that $n \ge 13$.
We prove that, for any coloring with two colors of the elements of the set $\{2, 3, \dots, 13\}$, the equation $x + y = z$ has monochrome solutions.
Assume the contrary: there exists a coloring such that there is no monochrome solution to the equation.
Case 1: If $2$, $3$, $4$ have color 1, then $5 = 2+3$, $6 = 2+4$, $7 = 3+4$ need to have color 2, hence $11 = 5+6$, $12 = 5+7$, $13 = 6+7$ need to be of color 1. But then $2+11 = 13$, and the equation has a monochrome solution (of color 1).

Case 2: If $2$ and $3$ are of color 1 while $4$ has color 2, then $5 = 2+3$ has color 2, and $9 = 4+5$ has color 1. But $3+6 = 9$, and numbers $3$ and $9$ have color 1, hence $6$ needs to be of color 2. Similarly, as $2+7 = 9$ and $2$, $9$ have color 1, it follows that $7$ has color 2. If $11$ has color 1, then $2+9 = 11$ leads to a monochrome solution. If $11$ has color 2, then $5+6 = 11$ is a monochrome solution.

Case 3: If $2$ and $4$ have color 1 while $3$ has color 2, then $6$ has color 2, $9$ has color 1. As $4$ and $9$ have color 1, we need $5$ to have color 2. Similarly, $2$ and $9$ are of color 1, therefore $7$ has color 2, $8 = 3+5$ has color 1. But $12 = 4+8 = 5+7$, hence we have a monochrome solution (of color 1 or color 2, depending on the color of $12$).

Case 4: $2$ has color 1, $3$ and $4$ have color 2. Then $7$ has color 1. But $2$ and $7$ having color 1 means that $5$ needs to be of color 2. This leads to $9 = 2+7 = 4+5$ and, again, regardless on the color of $9$, the equation has a monochrome solution.