Limit on how many photons can hit a surface As I understand it, photons have no mass and thus have no "dimensions". So I would expect there is no limit at how many photons can hit, say, a square meter of a given surface. But I am not sure, so -
Is there any theoretical limit on how many photons can hit a given unit of surface?
 A: I think you're asking how close photons can get to each other. i.e. if they were like 1 cm marbles then only one marble per sq cm could fit on a surface.  They are not truly dimensionless, but they are waves, and if you look up Bose–Einstein statistics https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics you'll learn that they do not repel each other like electrons etc tend to do, so they can be put very close together. Laser light hitting a surface is a good example of many photons in a small area.
One factor is probably the density of the surface material. When photons impact a surface they are typically being absorbed by electrons, and there could theoretically be too few electrons to absorb a very dense stream of photons. Just as likely, imagine a semi-porous material where some of the photons just go right through. To maximize absorption you also need a dense and thick material. Other factors could affect the answer you're looking for, like the reflectivity of the surface, for instance.
A: There is generally no hard limit to the number of photons per unit area in the sense that photons can overlap without issue (i.e., don't think of them as hard billiard balls). This ultimately comes about because photons are bosons, so the Pauli exclusion principle does not place restrictions on them like it does for fermions.
However, there can be many practical limits to the number of photons per unit area. These will depend on the frequency of the photons, as it may be easier to have lots of low-frequency RF photons compared to lots of gamma ray photons. One such limit is the Schwinger limit, where photons start to interact with each other and thinking of them as independent particles does not work anymore. This limit arises when the total electric field of the photons is large enough to create electron-positron pairs out of the vacuum.
If you have enough photons to reach the Schwinger limit, then you reach a non-linear regime, where some of those photons will (nearly) instantly disappear and turn into electron-positron pairs (and eventually back to photons), and the remaining photons will not behave at all like the light you are familiar with, so they should really be called something else - perhaps a "photon liquid"? Maxwell's equations don't apply here as in linear optics, so the physics is truly different.
The Schwinger limit is when the electric field reaches
$$E_s = \frac{(m_e c^2)^2}{e \hbar c}$$
The electric field from a source with $N$ photons per second with energy $\hbar \omega$ per photon over an area $A$ is given by
$$E=\sqrt{\frac{2 N \hbar \omega}{c \epsilon_0 A}}$$
Equating the two and solving for $N/A$ gives
$$N/A=\mathrm{Photons/Area}=\frac{1}{\hbar \omega} (\frac{(m_e c^2)^2}{e \hbar c})^2$$
$$\mathrm{Photons/cm^2}_{\mathrm{Schwinger}} = \lambda [nm] \times 2.33\cdot10^{45} \frac{\mathrm{Photons}}{\mathrm{Second\cdot cm^2}}$$
where $\lambda$ is the photon wavelength in nanometers.
This limit is extraordinarily high, and has not been achieved on earth with any man-made light source yet.
Another practical limit is the number of photons needed to destroy any material via ionization. Here you could perhaps use the Rydberg constant ($13.6$ Volts) and Bohr radius ($\sim 50$ picometers)  to construct an estimate for an electric field that would rip apart electrons from any atom. This estimate gives
$$\mathrm{Photons/cm^2}_{\mathrm{ionization}} = \lambda [nm] \times 10^{32} \frac{\mathrm{Photons}}{\mathrm{Second\cdot cm^2}}$$
A: My answer would be rather based on absolute theoretical limit. Schwinger limit is nice, but given that in case of non-linear effects some of the photons will be converted into electron-positron pairs,- one can still try to increase photon amount per unit area, conserving total energy per unit area in the form of $~e^-~+~e^+ ~+~ \gamma$.
Besides at least some electron-positron pairs will annihilate back into photons, so we will have concurrent processes. And even more,- vacuum has limited energy density. Which means that you can't generate as many electron-positron pairs as you wish.
Anyway, based on Planck units absolute theoretical energy density limit is :
$$ \Phi(J/cm^2) = m_{_P} c^2 ~\ell_{_P}^{−2} \tag 1,$$
which is about $\approx 10^{74}~\text{J/cm}^2$.
Converting (1) as number of photons :
$$ \Phi(n/cm^2) = \frac {m_{_P} c^2}{h~\nu} ~\ell_{_P}^{−2} \tag 2,$$
Taking as a reference visible light spectrum, for example when $\nu = 600~ \text{THz}$, from (2) we get that for visible photons this absolute energy density limit is $\approx 10^{93}~\text{photons/cm}^2$, or else vacuum break-down doesn't make sense.
