I have been fiddling around with this question and a definite proof would highly be appreciated.
My reasoning in a nutshell is
Due to the nature of irrationals; cannot be simplified, this means that they are not divisible by any other number with exception to 1 and self which means they must be primes.
Prime numbers are by definition integers with certain special properties. An irrational number is not an integer, and so cannot be a prime number.
To give some explanation for why only integers are defined to be prime numbers, note that the concept of divisibility is highly sensitive to what kind of numbers you allow yourself to use. For instance, $3$ is not divisible by $2$ as an integer: there is no integer $x$ such that $2x=3$. However, $3$ is divisible by $2$ as a real number, since there does exist a real number $x$ such that $2x=3$. So our usual notions of divisibility only behave the way we would expect when we restrict our attention to integers.
(More generally, you can define "prime" or "irreducible" elements in any integral domain which generalize the notion of "prime numbers" to other settings. But "prime number" always refers to these notions in the setting of the integers, and in any case there is not any natural setting in which all irrational numbers would be prime or irreducible. In the real numbers, for instance, every nonzero element is a "unit" and is not considered prime, similar to how $1$ is not considered a prime number.)
Finally, let me remark that I don't know what you mean when you say irrational numbers "cannot be simplified". The definition of an irrational number is a real number which cannot be written as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers. This doesn't particularly have anything to do with "simplification", whatever you might mean by that, nor does it have anything to do with divisibility. For instance, the number $2\pi$ is irrational: would you say it is divisible by $\pi$? Would you say it is divisible by $2$? In fact, it is reasonable to say it is divisible by every nonzero real number, if you are working in the context of real numbers, since if $d$ is a nonzero real number then $2\pi/d$ is also a real number.