That's the sum of two squares, since it can be written:

(2y)² + (3z²)²

Normally you can't factor the sum of two squares.  

----------------------------------------------

In rare cases you can but this time you can't.

-----------------

If your problem had been

y4 + 4z4

instead you could factor it by adding and subtracting 4y²z²

y4 + 4z4 + 4y²z² - 4y²z²

Rearranging terms:

y4 + 4y²z² + 4z4 - 4y²z²
 
(y4 + 4y²z² + 4z4) - 4y²z²

(y² + 2z²)² - 4y²z²

[(y² + 2z²) - 2yz][(y² + 2z²) + 2yz]

or

(y² - 2yz + 2z²)(y² + 2yz + 2z²)

But that's an advanced factoring technique, and will
not work on the problem you had without using square
roots.  So the answer to your problem is "It is a prime polynomial".

Edwin