Time dilation has nothing to do with how often photons are emitted or detected. It doesn't matter if the second photon was detected in a negligible amount of time or instantaneously from the first photon. Furthermore, it doesn't matter if the passage of time occurs in increments of Planck time. When considering time dilation in special relativity, the only things that matters are the constancy of the speed of light, the relative speed of the object being observed in the chosen coordinate system, and that the passage of time occurs.
Look, I understand that you have an idea about what time "actually" is but, before you make claims that time dilation is busted, you really need to understand what time dilation is and how to derive the equations mathematically. Otherwise, you end up with a bunch of statements that make no sense at all. First, we need to agree on a few things before we can actually understand what scientists are referring to when they talk about time dilation.
The passage of time occurs. How it occurs is not important; only that is does.
We can always choose a coordinate system where the object being observed changes position through space.
The speed of light is constant regardless of the chosen coordinate system.
Now, let's consider two identical light clocks where the reflector plates are separated exactly one-half the distance defined by the Planck length. Light clocks measure time by "bouncing" a photon between two reflectors. Because the speed of light is constant, it will take exactly one Plank time for the light to go from the bottom plate to the top plate and from the top plate back to the bottom plate when the clocks are at rest relative to the observer. Figure 1 illustrates the path that photons traverse when the light clock is at rest relative to the observer.
Figure 1: A light clock at rest relative to the observer.
Let [math]\mathcal{L}[/math] equal the distance between the reflective plates of the light clock. Because photons traverse the distance between the reflective plates at the speed of light, the total amount of time it takes for photons to go up and back down is equal to
[math]\Delta t=\frac{2\mathcal{L}}{c}=\frac{\sqrt{\frac{\hbar G}{c^3}}}{\sqrt{c^2}}=\sqrt{\frac{\hbar G}{c^5}}[/math].
As we can see, when the light clock is at rest relative to the observer, each tick of the clock occurs at intervals of Planck time. So far, everything is easy to understand, and we've managed to capture part of your idea that time occurs at discreet intervals of Plank time as defined by how our light clocks actually measure time. Now, let's consider a light clock that is moving with a constant speed relative to the observer.
Figure 2: A light clock in motion relative to the observer.
As illustrated by Figure 2, the distance light has to traverse between the two plates has increased because the clock is in motion relative to the observer. Photons can no longer move up and down and remain between the reflective plates of the light clock in motion. The path the photons traverse must also consider that the plates have moved through space with a speed of [math]v[/math] for every interval of time measured. Because we know how fast the clock is moving relative to the observer, the distance the clock has moved for each interval of time is equal to [math]v\Delta \tau[/math]. Using the distance between the plates and the distance the clock has moved for each interval of time, we can apply the Pythagorean theorem to calculate the total distance, [math]\mathcal{R}[/math], photons must traverse between each plates.
[math]\mathcal{R}=\sqrt{\left(\frac{1}{2}v\Delta \tau\right)^2+\left(\mathcal{L}\right)^2}[/math]
The total distance the photons have to traverse for each tick of the clock that is in motion as measured by the observer at rest is equal to [math]2\mathcal{R}[/math]. However, the speed of light is constant. So, the total amount of time the photons take to bounce between the plates is equal to
[math]\Delta \tau=\frac{2\mathcal{R}}{c}[/math],
[math]\Delta \tau=\frac{2\sqrt{\left(\frac{1}{2}v\Delta \tau\right)^2+\left(\mathcal{L}\right)^2}}{c}[/math].
Solving for [math]\Delta \tau[/math] we get:
[math]\Delta \tau^2=\left(\frac{2\sqrt{\left(\frac{1}{2}v\Delta \tau\right)^2+\left(\mathcal{L}\right)^2}}{c}\right)^2=\frac{v^2\Delta\tau^2+4\mathcal{L}^2}{c^2}[/math]
Multiply both sides by [math]c^2[/math]:
[math]c^2\Delta \tau^2=v^2\Delta\tau^2+4\mathcal{L}^2[/math]
Move the [math]\Delta \tau[/math] terms to the left side:
[math]c^2\Delta \tau^2-v^2\Delta\tau^2=4\mathcal{L}^2[/math]
Factor out [math]\Delta \tau^2[/math] from the left side:
[math]\Delta \tau^2\left(c^2-v^2\right)=4\mathcal{L}^2[/math]
Divide both sides by [math]c^2-v^2[/math]:
[math]\Delta \tau^2=\frac{4\mathcal{L}^2}{c^2-v^2}[/math]
Factor out [math]4\mathcal{L}^2/c^2[/math] from the right side:
[math]\Delta \tau^2=\frac{4\mathcal{L}^2}{c^2}\frac{1}{1-\frac{v^2}{c^2}}[/math]
Take the square root of both sides:
[math]\Delta \tau=\frac{2\mathcal{L}}{c}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]
Because [math]\Delta t=2\mathcal{L}/c[/math], we can substitute in [math]\Delta t[/math] and arrive at the equation for time dilation in special relativity:
[math]\Delta \tau=\frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}[/math]
This is a mathematical definition for what time dilation is according to special relativity, and our experiments confirm that this form of time dilation does, in fact, exist! Pretty cool huh?
As you can see, time dilation doesn't depend on any of these crazy notions you keep repeating, and relativity didn't change.