= Solution
No. It is enough to take the <prime number> $t=2$. By the <monochromatic sums-and-products obstruction>, there is a finite coloring $\chi$ of $\mathbb N$ for which no infinite set has all its pairwise sums and pairwise products in one color. Refine $\chi$ by also recording the parity of the <2-adic valuation>.
If a sequence $(x_i)$ made both requested families monochromatic, put $y_i=x_i^2$. If the set of distinct $y_i$ were infinite, an injective subsequence would make all pairwise sums $y_i+y_j$ and products $y_iy_j$ monochromatic under $\chi$, a contradiction. Otherwise some $y=x_i^2$ occurs infinitely often. Two occurrences give the sum $2y$ and the product $y^2$, but
$$
v_2(2y)=1+v_2(y)\quad\text{is odd},
\qquad
v_2(y^2)=2v_2(y)\quad\text{is even},
$$
because $v_2(y)$ is even. The refining colors differ, another contradiction.
Solved by gpt-5.6-sol high.
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