What does happen if I push a piston in an adiabatic cylinder containing an ideal gas?

During an adiabatic process, $pV^gamma=const$, where $gamma$ is the adiabatic factor. Using the ideal gas equation, you can calculate the changes in all three parameters (namely $p,V,$ and $T$). Your prediction is correct, for most gasses pressure and temperature will increase, and volume will decrease, which can be seen in a $(p,V)$ diagram from the flow of an adiabatic line.

Without a mathematical analysis (leading to the equation given by Soba noodles), it's possible to make a qualitative prediction… We know from $PV=nRT$ that in an isothermal compression (T constant) the pressure rises, with p proportional to $frac {1}{V}$. But in an adiabatic compression, T rises, because the work done on the gas raises its internal energy. So, appealing again to $PV=nRT$, both the decrease in V and the increase in T contribute to the rise in P, which is therefore steeper than in the isothermal case.

What does happen if I push a piston in an adiabatic cylinder containing an ideal gas? Is it possible to predict what's going to happen? Will Temperature and Pressure increase while Volume decreases?

If you push a piston in an adiabatic cylinder (and piston) you will do work compressing the gas. This will reduce the volume of the gas and increase both the temperature and pressure of the gas. These can be predicted as follows:

FINAL TEMPERATURE

Since the process is adiabatic, $Q=0$, and since $Delta U=Q-W$ then $Delta U=-W$. $W$ is negative since the work is done on (energy into) the gas. Consequently, there will be an increase in internal energy of the gas. Since the internal energy of an ideal gas is a function of temperature only, you can predict there will be an increase in temperature.

FINAL PRESSURE

In addition to the ideal gas equation, which defines the relationship between the variables at equilibrium (the initial and final states of any process involving an ideal gas), an adiabatic process obeys the following relationship:

$$PV^{k}= constant$$ Where $k=frac {C_p}{C_v}$

Or

$$frac{P_2}{P_1}=Biggl(frac {V_1}{V_2}Biggr)^k$$

From the last equation we can see that since the final volume, $V_2$ is less than the initial volume, $V_1$, then you can predict that the final pressure $P_2$ will be greater than $P_1$.

You can calculate the final temperature and pressure if you know the following:

1. The initial pressure, volume, temperature and the number of moles or mass of the gas.
2. The final volume after completing the compression.
3. The specific heats ($C_p$, $C_v$) for the gas.

Do we have to control one Variable in all processes?

To answer this you would need to clarify what you mean by “controlling” and by “variables”.

Hope this helps.