Can photons decay without interaction? Can photons decay like other particles without interacting with other particles or fields, i.e. by just "being"?
In case the answer is "no" - does this have anything to do with them travelling at c, so their clocks as seen from our frame of reference is standing still?
 A: The answer is "no", not as a straight 1-body -> 2-body decay.  Proof:  let $E_\gamma, E_1$, and $E_2$ denote the energies of the initial photon and the two daughters, and define $\vec{p}_\gamma$, $\vec{p}_1$, and $\vec{p}_2$ similarly for their momenta.  By conservation of energy and momentum, we have
$$
E_\gamma = E_1 + E_2,
$$
$$
\vec{p}_\gamma = \vec{p}_1 + \vec{p}_2.
$$
Now we apply a bunch of inequalities.  First, we note that $|\vec{A} + \vec{B}| \leq |\vec{A}| + |\vec{B}|$, with equality only holding when the two vectors are parallel.  Thus, we have
$$
p_\gamma = |\vec{p}_1 + \vec{p}_2| \leq p_1 + p_2
$$
(where we're denoting the magnitudes of the momenta without the arrow.)  Next, we note that for any particle, $c p = \sqrt{E^2 - m^2} \leq E$, with equality holding only when the particle is massless.  This means that
$$
p_1 + p_2 \leq \frac{E_1}{c} + \frac{E_2}{c}.
$$ 
Finally, we note that $p_\gamma = E_\gamma/c = (E_1 + E_2)/c$.  Putting these all together, we end up with
$$
p_\gamma \leq p_\gamma,
$$
with equality holding only when the two daughter particles are (a) traveling in the same direction, and (b) massless.  This one possible case is something of an "edge case";  since there is only one possible state that the photon can decay to, out of all the possible momenta that the daughter particles could have, it makes it nearly impossible for the photon to decay this way.  (In the parlance of particle physics, the allowed volume of phase space for the daughter particles is effectively zero.)
As far as the second question, of whether this can be viewed as a product of the $v \to c$ limit of relativistic time dilation:  I don't think so.  The approach you're advocating would kind of work as follows:


*

*Model the photon as a particle with mass $m_\gamma$, and a mean lifetime $\tau_0$ in its rest frame.

*Calculate the lifetime of a photon with energy $E$, still assuming $m_\gamma \neq 0$.  The Lorentz factor of this photon would just be $\gamma = E/m_\gamma c^2$, and so the mean lifetime would be $\tau(E) = E \tau_0 / m_\gamma c^2$.

*Set $m_\gamma \to 0$ while holding $E$ constant.  Hey presto, $\tau$ is now infinite.


The problem with this approach is that it doesn't capture the fine details of the above proof:  in the above calculation, we saw that the main impediments to the decay are (a) the mass of the daughters and (b) the alignments of their momenta.  It's not just that the photon has no time passing;  it's that even if it wanted to decay, there's no possible two-particle state for it to decay to.
A: I found a nice discussion of this from several years ago at physicsforums. Inside, Ben Crowell makes an argument that does in fact use time dilation: if you're willing to accept that the properties of a massless photon are the same as the $m\rightarrow 0$ limit of a massive photon, it's easy to see the decay rate has to be zero. One way to see this is that as you take $m \rightarrow 0$ (and thus $v \rightarrow c$), the decay rate goes to 0 in any frame that is not the particle's rest frame, which of course ceases to exist at that limit. Since the massless photon propagator in QED can be found by taking a m=0 limit of a massive propagator, this assumptions seems pretty reasonable to me, but maybe someone who knows more QFT than I could give a stronger opinion.
