= Solution
Write $\phi_B=Z^{1/2}\phi$, $Z=1+\delta_Z$, and define the <mass counterterm> $\delta_m$ as a squared-mass coefficient:
$$
\mathcal L_{\rm ct}=\frac12\delta_Z(\partial\phi)^2-\frac12\delta_m\phi^2,
\qquad \delta_m=Zm_B^2-m^2.
$$
With the <self-energy> convention in <1/b>, the inverse <scalar propagator> is $p^2-m^2+\delta_Zp^2-\delta_m-\Sigma(p^2)$. Requiring the <pole mass> to equal $m$ therefore gives the <on-shell renormalization scheme> condition
$$
\delta_m=m^2\delta_Z-\Sigma(m^2).
$$
At one <loop order>, $\Sigma_1'=0$; hence a unit <pole residue> also gives $\delta_Z^{(1)}=0$, and
$$
\boxed{\delta_m^{(1)}=-\frac{g_4}{8\pi}\log\left(1+\frac{\Lambda^2}{m^2}\right),\qquad
m_B^2=m^2+\delta_m^{(1)}+O(g_4^2).}
$$
The <mass counterterm> <Feynman rule> is $-i\delta_m$, cancelling the <tadpole diagram>. A convention defining a positive mass subtraction by $m_B^2=m^2-\delta_m^{\rm sub}$ instead has $\delta_m^{\rm sub}=-\delta_m$; specifying this convention avoids a spurious sign disagreement.
There is no one-loop <wave-function renormalization> because the <tadpole diagram> is independent of <momentum>. Each one-loop four-point <bubble diagram> has two <scalar propagators>, so its large Euclidean <loop momentum> integral behaves as $\int d^2k_E/k_E^4$ and is <ultraviolet finite>. Thus no divergent <quartic interaction> <counterterm> is required. Finite coupling redefinitions remain possible. At higher <loop orders>, finite <wave-function renormalization> may be needed if unit <pole residue> is imposed; <power counting> only rules out its <ultraviolet divergence>.
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