AES: Given two deterministic plaintexts, compute two keys let the ciphertext is the same, is it possible?
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.