The way that you ask your question confuses the answer because you say "Will the speed of universe expansion will make the sky bright but the red shift make it invisible to our eyes", because the sky is already "bright" in certain wavelengths, particularly the centimeter cosmic background radiation (CMB).
Other than this revision, yes, the observation that the night sky is dark has been a clear argument against an infinitely old universe since long ago. The evidence for the big bang in the form of a consistently increasing red shift pretty much seals the deal for myself, regarding the fact that the universe has an age.
Furthermore, over time, you are entirely correct in the assertion that the number of observable objects will increase drastically, and quite possibly infinitely. Consider that we only see $x$ distance away which terminates at the CMB, thus limiting the number of galaxies we can see, with the furthest galaxies being the earliest evolutionary stage of galaxies. The number of "young" galaxies we can see will progressively increase as more of the veil from the CMB is pulled back through the arrival of the new light. The "young" galaxies we can see now will mature and the total number will increase. Whether or not this will increase forever is disputable since dark energy pulls space apart could prevent it but we can't claim to know exactly what the behavior of dark energy far in to the future will be.
Additions
I started thinking about the problem more and I wanted to formalize things a bit better. Take the most basic case, we'll deal with a flat Newtonian space for now. As before, take $x$ to be the distance to a certain galaxy we are current seeing. Take the present time (after the big bang) to be $t$ and that we're observing that galaxy at $t'$. It follows...
$$x=c (t-t')$$
Imagine the universe has a galaxy density of $\rho$ galaxies per unit volume. Then knowing that, we can actually write the rate $r$ at which galaxies older than $t'$ are appearing into our view. This is done knowing the surface of a sphere is $4\pi r^2$.
$$r=4 \rho c \pi x^2$$
It's fascinating to consider that in a line connecting every object in the night sky and us, there exists the entire history of the object encoded in the light waves making their way to us. One way to talk about the acceleration of the universe is to say that there is a slowdown in the rate at which we are receiving this information. We are watching the far off objects in slow motion.
If we make the obviously incorrect but useful assumption that all objects emit light at the same rate at all times, then the intensity we see will be proportional to $1/x^2$, and given some $S$ which is, say, the number of photons emitted total per unit time, then the intensity of light we receive from a given body would be $S/(4 \pi x^2)$. Multiplying this by the rate, we can get a very nice equation for $s(x)$ which is the contribution to the number of photons we receive from the differential "shell" of stars at $x$.
$$s(x) = S \rho c $$
This equation is important because it is cumulative from time at $t'$ to $t$, meaning that the objects that entered our field of vision from the "genesis" of that type of object are still contributing to the population of photons reaching us today. So the number of photons we are receiving could be said to be:
$$\int_{t'}^t S \rho c dt = S \rho c (t-t')$$
A more advanced view of the situation simply notes that the "movie" for each of these stars is being played in slow motion. We'll just define a factor for that and put it in the equation.
$$l(x) = \frac{\Delta t_{object}}{\Delta t_{Earth}}$$
I should preface this by saying that this isn't actually saying time is going slower for that object, and this isn't even the time dilation as defined by general relativity, this is the time dilation you would measure by watching a clock in a galaxy far away with a space telescope and comparing it to the local time. Yes, these two are different, and yes, I am avoiding advanced relativistic concepts by making it an accounting problem. Now the total # of photons we receive per unit time is the following.
$$\int_{t'}^t S \rho c l(x) dt$$
I won't use any calculus chain rules because there's no guarantee that $l(t)$ is any more helpful to you than $l(x)$! But I should also note that the final $x$ you get in this equation at $t$ will be meaningless. It is not the general relativity distance, it's some bastardization of that by using $c t$, which is clearly not how it actually works. Nonetheless, there is some usefulness in the above equation. We can even identify the radiative energy being received by considering the energy of the photon being proportional to it's frequency, with $E_e$ being the energy of the emitted photon and $E_o$ the observed photon.
$$\frac{E_e}{E_o} = l(x)$$
And the total energy would then be the following with $h$ the familiar plank constant.
$$E = \int_{t'}^t S h \rho c l(x)^2 dt$$
Anyway, my intent is for these to be instructive "kindergarten" equations for the subject. The bottom line is still clear from them - that the # of photons reaching us would increase linearly over time but it's less since $l(x)\le 1$. Similarly, the radiative energy reaching us would be less by even a smaller factor due to the redshift. I hope this is a clear picture.