= Solution
First take the <trace> in four <spacetime> dimensions:
$$
T^\mu{}_{\mu}=-(\partial\phi)^2+2m^2\phi^2+\frac{\lambda}{6}\phi^4,
\qquad
\Omega^\mu{}_{\mu}=T^\mu{}_{\mu}-3C\Box\phi^2.
$$
The <product rule> gives $\Box\phi^2=2(\partial\phi)^2+2\phi\Box\phi$. For the massless <equation of motion>, $\phi\Box\phi=-\lambda\phi^4/6$, so
$$
\left.\Omega^\mu{}_{\mu}\right|_{F=0,m=0}
=-(1+6C)(\partial\phi)^2+\lambda\left(\frac16+C\right)\phi^4.
$$
Both coefficients vanish for
$$
\boxed{C=-\frac16.}
$$
The minus sign follows from the order $\partial^\mu\partial^\nu-\eta^{\mu\nu}\Box$ used in the question's <stress-energy tensor improvement>. Commuting <partial derivatives> shows that this improvement has identically zero <divergence>, so it preserves the conservation law.
The precise <trace> identity, without using the <equation of motion>, is
$$
\boxed{\Omega^\mu{}_{\mu}=m^2\phi^2+\phi F.}
$$
Hence the massless <tensor> is traceless \b[on shell]. It is not traceless for arbitrary <off shell> configurations: a nonzero constant $\phi=A$ with $\lambda\ne0$ has $\Omega^\mu{}_{\mu}=\lambda A^4/6$. This is a classical result; it does not assert absence of a quantum <trace anomaly>.
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