Let $A$ and $B$ be events in the probability space $\Omega$.
Two events are mutually exclusive iff $A\cap B = \emptyset$
In simpler terms, they cannot occur simultaneously. We have the useful result that $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$ in general, which in the case of mutually exclusive events will simplify as $Pr(A\cup B) = Pr(A)+Pr(B)$ since $Pr(\emptyset)=0$.
An example of mutually exclusive events would be "flipping heads on a coin" versus "flipping tails on a coin" in a single coin toss.
Two events are independent iff any of the following are true:
$$Pr(A\cap B) = Pr(A)Pr(B)\\
In simpler terms, independent events have no influence on one another in terms of occurring. For example, if you flip a coin and roll a six-sided die at the same time, the event "the coin-flip is a heads" is independent to the event "the die face is showing a four"
Suppose that $A$ and $B$ are simultaneously mutually exclusive and independent.
Then we have $Pr(A\cap B) = Pr(\emptyset)=0$ because of mutual exclusivity, while we also have $Pr(A\cap B) = Pr(A)Pr(B)$ because of independence.
This implies $Pr(A)Pr(B)=0$.
Now, two real numbers multiplied together equals zero if and only if at least one of the numbers is zero. That implies $Pr(A)=0$ or $Pr(B)=0$. Thus, at least one of $A$ or $B$ is an (almost) impossible event (an event of probability zero).
By contraposition, if $A$ and $B$ are both events of positive probability, then $A$ and $B$ cannot simultaneously be mutually exclusive and independent. It is possible that they are one or the other or neither, but not both.