Suppose #a^2+b^3=c^4# for some prime numbers #a, b, c#.
Note that #a=b=c=2# does not work, meaning for both sides of the equation to have the same parity, one of #a, b, c# must be #2#, and the other two must be odd primes.
Next, note that we may subtract #a^2# from each side of the equation to get
#b^3 = c^4-a^2 = (c^2+a)(c^2-a)#
#=> b^2 = c^2+a# and #b = c^2-a#
(as #c^2+a > c^2-a# and #b# is prime)
#=> (c^2-a)^2 = c^2+a#
We now consider three cases.
Case 1: #a = 2#
#=> (c^2-2)^2 = c^2+2#
#=> c^4-4c^2+4 = c^2+2#
#=> c^4-5c^2+2 = 0#
#=> c^2 = (5+-sqrt(17))/2#
#=> c !in NN#
but this contradicts the premise that #c# is a prime.
Case 2: #b = 2#
#=> {(c^2-a = 2),(c^2+a = 4):}#
#=> (c^2-a)+(c^2+a) = 2+4#
#=> 2c^2 = 6#
#=> c^2 = 3#
#=> c != NN#
again contradicting the premise that #c# is prime.
Case 3: #c = 2#
#=> a^2+b^3=16#
If #a# and #b# are both odd primes, then the least the left hand side can attain is when both #a# and #b# are the least odd prime, i.e. #a=b=3#, giving #a^2+b^3 >= 3^2+3^3 > 16#, a contradiction.
As each case leads to a contradiction, there are no three primes #a, b, c# satisfying #a^2+b^3=c^4#
