Does the equation $$2006 = (x+y)^2 + 3x + y$$ have a solution in nonnegative integers?

Yes. Put $$s = x + y$$; then $$3x + y = s + 2x$$ and the equation becomes $$s^2 + s + 2x = 2006$$ with $$0 \le x \le s$$. This forces $$s(s+1) \le 2006 \le s^2 + 3s$$, which holds only for $$s = 44$$ (since $$44 \times 45 = 1980$$ and $$44^2 + 3 \times 44 = 2068$$). Then $$2x = 2006 - 1980 = 26$$, so $$x = 13$$ and $$y = 31$$. Check $$(13 + 31)^2 + 39 + 31 = 1936 + 70 = 2006$$, so $$(x, y) = (13, 31)$$ is a solution, and it is the only one.