= Solution
Take $M^2>0$ initially to avoid an infrared divergence. A <Schwinger parameterization> and the Euclidean <Gaussian integral> give, in the convergent range $0<\operatorname{Re}d<4$,
$$
\frac1{(k^2+M^2)^2}=\int_0^\infty dt\,t e^{-t(k^2+M^2)},\qquad \int\frac{d^dk}{(2\pi)^d}e^{-tk^2}=(4\pi t)^{-d/2}.
$$
Integration over $t$ supplies the <Gamma function>. Its <analytic continuation> defines the answer in <dimensional regularization>:
$$
\boxed{I_d(M^2)=\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}(M^2)^{d/2-2}.}
$$
Use $d=4-\epsilon$ near four dimensions and $d=6-\epsilon$ near six dimensions. The <Gamma function recurrence> implies $\Gamma(\epsilon/2)=2/\epsilon+O(1)$ and $\Gamma(-1+\epsilon/2)=-2/\epsilon+O(1)$. Therefore
$$
\boxed{I_{4-\epsilon}(M^2)=\frac{2}{(4\pi)^2\epsilon}+O(1),\qquad I_{6-\epsilon}(M^2)=-\frac{2M^2}{(4\pi)^3\epsilon}+O(1).}
$$
Both are simple poles. Equivalently their residues as functions of $d$ are $-2/(4\pi)^2$ at $d=4$ and $2M^2/(4\pi)^3$ at $d=6$. The negative pole near six dimensions is a feature of <analytic continuation>; it is not a negative value of the original convergent positive integral. A massless zero-momentum integral is scaleless and cannot be used indiscriminately to extract these ultraviolet poles, because <dimensional regularization> then combines ultraviolet and infrared contributions.
For the following <Feynman diagrams>, specify the Euclidean <action> and additive <counterterms> explicitly so their signs are unambiguous. The four-dimensional interaction is $\mu^\epsilon\lambda\phi^4/4!$, the six-dimensional interaction is $\mu^{\epsilon/2}g\phi^3/3!$, and the common free inverse <scalar propagator> is $p^2+m^2$. Here $\epsilon=4-d$ or $6-d$, respectively. The factors of $\mu$ give dimensionless renormalized couplings and change no leading pole residues. The cubic Euclidean theory is treated as a formal perturbation expansion about its massive quadratic <scalar field theory>; its real cubic potential is not bounded below.
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