Here's an indirect proof that "photons" do bend spacetime. Consider the Peres metric (I'm using $c \equiv 1$ and the $(1, -1, -1, -1)$ convention):
\begin{equation}\tag{1}
ds^2 = dt^2 - dx^2 - dy^2 - dz^2 + F(x, y, t - z)(dt - dz)^2,
\end{equation}
where $F(x, y, u)$ is an arbitrary function of three independant variables ($u = t - z$). Substitute this metric into Einstein's equation. First: without any stress tensor (and no cosmological constant):
\begin{equation}\tag{2}
G_{\mu \nu} = 0.
\end{equation}
After some algebra, you then get a constraint on $F(x, y, u)$:
\begin{equation}\tag{3}
\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = 0.
\end{equation}
Thus, $F$ must be an harmonic function in $x$ and $y$. The simplest non-trivial solution (with spacetime curvature) is a linear superposition of quadratic functions (there are two independant polarization states for the gravitational wave):
\begin{equation}
F(x, y, u) = \mathcal{A}(u)(\, x^2 - y^2) + \mathcal{B}(u) \, x \, y,
\end{equation}
where $\mathcal{A}(u)$ and $\mathcal{B}(u)$ are abitrary functions of $u = t - z$. Metric (1) then describes a planar gravitational wave propagating in vacuum.
Then add a planar monochromatic electromagnetic wave, of energy-momentum
\begin{equation}\tag{4}
T_{\mu \nu} = \Phi(x, y, u) \, k_{\mu} \, k_{\nu},
\end{equation}
where $k^{\mu} = (\omega, 0, 0, \omega)$ is the wave number and $\Phi(x, y, u)$ is arbitrary. Einstein's equation then becomes
\begin{equation}\tag{5}
G_{\mu \nu} = -\, \kappa \, T_{\mu \nu}.
\end{equation}
Of course $\kappa \equiv 8 \pi G$. A lot of algebra gives the following constraint:
\begin{equation}\tag{6}
\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = 2 \kappa \omega^2 \, \Phi.
\end{equation}
I'm considering a planar monochromatic electromagnetic wave propagating in spacetime, with a circular polarization (this is a classical field which is the closest thing to a "quantum photon" of angular frequency $\omega$):
\begin{equation}\tag{7}
A^{\mu}(x, y, u) = \varepsilon_1^{\mu} \, \mathcal{F}(u) + \varepsilon_2^{\mu} \, \mathcal{G}(u),
\end{equation}
where $\varepsilon_{1, \, 2}^{\mu}$ are the space-like polarization four-vectors, orthogonal to $k^{\mu}$, and
\begin{align}\tag{8}
\mathcal{F}(u) &= a_0 \cos{(\omega \, u)},
& \mathcal{G}(u) &= a_0 \sin{(\omega \, u)}.
\end{align}
The amplitude $a_0$ is just a constant. It is easy to verify that (7) and (8) give (4) with $\Phi(x, y, u) = \text{cste} \propto a_0^2$. Then (6) can be solved to give a simple non-trivial solution (the Riemann curvature tensor isn't 0):
\begin{equation}\tag{9}
F(x, y, u) = \frac{\kappa \, a_0^2 \, \omega^2}{8 \pi \alpha} \, (\, x^2 + y^2).
\end{equation}
($4 \pi \alpha$ is the electromagnetic coupling constant that appears in the energy-momentum tensor. It depends on your favorite units for the field amplitude $a_0$. I use the fine-structure constant $\alpha \approx \frac{1}{137}$). Metric (1) with the function (9) then describes a circularly polarized EM wave (and its associated gravitational wave) propagating in spacetime. The Riemann curvature isn't 0 (its components are constants, in that case, since the wave energy-momentum is homogeneous).
So the non-localisable "photon" do curves spacetime in a non-trivial way. Because of the circular polarization, curvature is homogeneous (but non-isotropic since the wave propagation defines a priviliged orientation).
