AES: Given two deterministic plaintexts, compute two keys let the ciphertext is the same, is it possible? Given two plaintexts $p_1$ and $p_2$, is it possible to find two keys for AES(-CBC) so that $$E_{k_1}(p_1) = E_{k_2}(p_2) $$ 

I want to make sure that will never happen, or is there any way to prevent that from happening? Because that will make my system unreliable.
What is the mathematical property stating that it is hard to find a collision in the AES algorithm?

The answer of Lindell is important.

For an adversary with infinite computation power it is always possible to find $p_1$ and $p_2$ s.t. $E_{k_1}(p_1) = E_{k_2}(p_2)$. Choose any values $k_1, k_2$ with $k_1 \neq k_2$ and calculate $p_3 = E^{-1}_{k_2}(E_{k_1}(p_1))$ and stop if $p_3 = p_2$, otherwise repeat with different $k_2$.

However if the question is how easy it is to find such $k_1, k_2$ it is a different story.