If we color $5$ consecutive integers in $2$ colors, then there must be a 3-term arithmetic progression with first two elements of the same color.
Indeed, if the first two elements are of the same color then we are done, and likewise we must have that the first four elements alternate colors, and then we have no color available for the fifth element. $4$ integers do not suffice, by the sequence of colors
RGRG. We can ask:
How many consecutive numbers are needed to ensure that any $m$-coloring contains a 3-term arithmetic progression whose first two elements are monochromatic?
For $m=1$ it is $3$.
For $m=3$, by trial and error, I think $15$ elements do it. $14$ do not suffice, by this sequence of length 14 (which cannot be extended):
RGRG BRBR GBGB RR.
For general $m$ we can no more just determine the next value based on the previous ones.