How to explain that an increase in temperature favours an endothermic reaction?
I believe that collision theory in itself wouldn't be enough to explain this behavior. You can explain it by having Arrhenius equation and the activation energy in mind. What you got wrong was the effect of temperature on reaction rates depending on the activation energies. Actually, the speed constant of a reaction with higher activation energy will be more significantly affected by variation in temperature. This can be verified easily by manipulation of the Arrhenius equation. Apply ln to both sides to obtain:

$\ln{k}=\ln{A}-\frac{E_a}{RT}$

The graph of ln(k) as a function of (1/T) is a straight line with slope's magnitude directly proportional to activation energy.

Now, assuming that the endothermic reaction has higher activation energy than its inverse, it will be more affected by, say, an increase in temperature. While both reactions will become faster, the speed constant of the endothermic reaction will increase to a greater degree and that will shift the equilibrium favoring the products of the endothermic reaction.
If the temperature is decreased, the endothermic reaction will be slowed down to a greater degree, favoring the products of the exothermic reaction.