I think you're either misusing the word "renormalizable" or using it in a very old-fashioned way. Perturbative quantum gravity is renormalizable at all loop orders in the sense that divergences can be removed by the addition of local counterterms. Like all EFTs for which the leading interaction is irrelevant (in this case the Einstein-Hilbert term) such counterterms are necessarily higher order in the derivative expansion. The claim about pure gravity is that it is *finite* at one-loop, meaning that there are no UV divergent one-loop amplitudes.

One can show that in \(d=4\) for \(\mathcal{N}\geq 1\)supersymmetry, pure supergravity is also finite at two-loops. Let's be clear what this means, it says that any two loop amplitude calculated with vertex factors from the supersymmetrization of the Einstein-Hilbert action is non-divergent if calculated with a supersymmetry preserving regulator. One way to show this is by a combination of counterterm analysis and on-shell methods. You begin by showing that the only possible counterterm would have to be a supersymmetrization of \((R_{\mu\nu\rho\sigma})^3\)and then show that no such supersymmetrization exists. A very efficient way to do the second step is to first see that such a counterterm would produce tree-level on-shell 3-point amplitudes for three gravitons with helicity assignments \((+2,+2,+2)\). However, such an amplitude can never satisfy the on-shell supersymmetry Ward identities and therefore cannot be present in a supersymmetric model.

Similar arguments can be extended to higher loop orders. The most interesting results have been obtained in \(\mathcal{N}\geq 5\) supergravity. At leading order in the derivative expansion all supergravities have an electromagnetic duality symmetry which acts linearly on the graviphotons and chiral fermions and non-linearly on the scalars (if present). At tree-level the non-linear symmetry gives rise to low-energy theorems which say that any amplitude vanishes in the limit that the momentum of a scalar is taken to zero. For \(\mathcal{N}<5\) there is a non-vanishing \((\text{Diff})^2(\text{EM-duality})\) anomaly which leads to a breakdown of the low-energy theorem at loop order. For \(\mathcal{N}\geq 5\) this anomaly vanishes and so we can regularize loop integrals in such a way that the duality symmetry is unbroken. Consequently the on-shell amplitudes arising from any potential counterterms must satisfy the vanishing scalar low-energy theorem.

For example, in https://arxiv.org/abs/1009.1643 it was shown explicitly in \(\mathcal{N}=8\) supergravity that the first on-shell amplitudes which satisfy \(\mathcal{N}=8\) supersymmetry, \(\text{SU}(8)_R\) symmetry and \(E_{7(7)}\) duality symmetry (importantly the low-energy theorems for the scalars) correspond to an interaction term of the form \(D^8R^4\) which in turn corresponds to a 7-loop counterterm. So we expect \(\mathcal{N}=8\) supergravity in \(d=4\) to be UV finite up to six-loops.