$φ^3$ theory: three-point, one-loop diagram ($D=4$)

First of all, how did you get the denominator? Are there some kinematic conditions on $p^2$ and $(p+q)^2$, which might be related to $q^2$ and $m^2$? Are you sure about the coefficient in front of $m^2$?

Nevertheless, I would show a way to proceed with your integral: begin{equation} I equiv int_0^1 dx int_0^1 dy int_0^1 dz frac{delta(1-x-y-z)}{x y q^2 + (1 - z + z^2) m^2}, end{equation} though I will not go into the detail that will be too cumbersome. After integrating over $z$ by using the delta function, and subsequently changing the variable $x to 1 - x$, you get begin{equation} I = int_0^1 dx int_0^x dy frac{1}{A x^2 + B y^2 + C xy + D x + E y + F} , end{equation} where begin{equation} A = B = F = m^2, qquad C = - (q^2+2m^2), qquad D = - m^2, qquad E = q^2 + m^2 . end{equation} This type of integrals appearing in one-loop calculations was thoroughly studied in the famous 1979 paper by 't Hooft and Veltman. The denominator is quadratic both in $x$ and $y$, which is rather tough to tackle, so the first step is applying a variable transformation in such a way that the denominator becomes linear in one of the variables. Then performing the integral with respect to the linear variable (with an adequate splitting of the integration domain and variable transformations) gives logarithms, whose arguments are now quadratic in the remaining variable. A quadratic argument can be factorised as a product of two linear terms, so the logarithms with quadratic arguments are split into logarithms with linear arguments. At the end of the day, integrating such logarithms (multiplied with some reciprocals) leads to dilogarithms (known as Spence's functions); indeed many terms with dilogarithms and logarithms in general.