= Solution
Set the external <four-momenta> to zero after identifying the proper diagrams. In each box or triangle, at least one endpoint of the internal <photon> is a cubic vertex adjacent to an external scalar of zero <momentum>. That vertex is proportional to $l_\rho$, whereas the supplied numerator obeys $l^\rho P_{\rho\sigma}(l)=0$. All seven such graphs vanish algebraically. This use of the transverse numerator is stronger than merely saying that one has chosen a generic <Lorenz gauge>.
The two surviving <seagull vertex> bubbles have vertex factors $2ie_D^2g^{\mu\rho}$ and $2ie_D^2g^{\sigma\nu}$. Combining their two <propagators> gives
$$
\boxed{i\Gamma_{\rm loop}^{\mu\nu}(0)
=-4e_D^4\int\frac{d^Dl}{(2\pi)^D}
\frac{g^{\mu\rho}P_{\rho\sigma}(l)g^{\sigma\nu}}
{(l^2+i0)(l^2-m^2+i0)}+(\mu\leftrightarrow\nu).}
$$
This <scalar-QED four-point vertex pole in transverse gauge> calculation determines the proportionality constant as well as the required tensor structure. Keep $m>0$ while extracting the <ultraviolet pole>. <Lorentz covariance> inside the regulated integral replaces $l^\mu l^\nu$ by $g^{\mu\nu}l^2/D$, giving the factor $(D-1)g^{\mu\nu}/D$.
For its scalar integral, <Feynman parameter> gives
$$
\begin{aligned}
I_D&=\int\frac{d^Dl}{(2\pi)^D}\frac1{(l^2+i0)(l^2-m^2+i0)}\\
&=\frac{i\Gamma(2-D/2)}{(4\pi)^{D/2}}
\int_0^1 dx\,[(1-x)m^2]^{D/2-2}\\
&=\frac{i\Gamma(2-D/2)(m^2)^{D/2-2}}
{(4\pi)^{D/2}(D/2-1)}
=\frac{i}{8\pi^2\epsilon}+O(1),\qquad D=4-\epsilon.
\end{aligned}
$$
There is no infrared divergence at $l=0$ with nonzero $m$. At pole order the two <photon> assignments and the factor $3/4$ therefore yield
$$
i\Gamma_{\rm loop}^{\mu\nu}\big|_{\rm pole}
=-\frac{3ie_D^2e^2}{4\pi^2\epsilon}g^{\mu\nu}.
$$
The common engineering scale factor is included in $e_D^2$; finite logarithms are immaterial to the pole. Adding $2ie_D^2\delta_4g^{\mu\nu}$ cancels it when
$$
\boxed{Z_4=1+\frac{3e^2}{8\pi^2\epsilon}+O(e^4)\quad(D=4-\epsilon).}
$$
This is <minimal subtraction scheme>, up to optional finite terms. It agrees with the transverse-gauge <scalar QED wave-function pole> and $Z_4=Z_2$. Setting $m=0$ at the outset would instead make the zero-momentum integral scaleless and obscure its separate ultraviolet divergence.
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