Using the infinitely many rational points on the curve $x^4-y^4=(2^4-1) y^2 z^2$, the smallest solution we found is with $x$ having 146 decimal digits (smaller solutions may exist).
Using the same curve, there are infinitely many solutions of very large size.
The solution in sage code:
def fourpowerful1():
"""
x^2,y^2,x^2+y^2,x^2-y^2 all powerful
Coming from the infinitely many points on the curve
x^4-y^4=(2^4-1) y^2 z^2
Author: Georgi Guninski, Thu Jul 4 10:21:11 AM UTC 2024
"""
x_1=51203929262112985953504884979884461019907289047719199306830900384193163023961621160885389604169689049920413633464220309380754803019095566545430404
y_1=5168374460623477253206993537636619974630340154166825903559471772312195920181455373026653181235106737210693085904835681664925203352727316123710721
tp=x_1+y_1
tm=x_1-y_1
print("gcd=",gcd(x_1,y_1),x_1.is_square(),y_1.is_square(),(tp//5^3).is_square(),(tm//3^3).is_square(),"digits(x)=",len(str(x_1)) )
print("x=",factor(x_1))
print("y=",factor(y_1))
print("x+y=",factor(tp))
print("x-y=",factor(tm))
I think there are infinitely many solutions via rational points on elliptic curves, but the smallest solution is large.
Let $X=x^2,Y=y^2$, then we want $(X^2-Y^2)*(X^2+Y^2)=X^4-Y^4 =\text{powerful}$
In this question Elkies gave solution to similar problem $X^4+Y^4=\text{powerful}$: $427511122^4 + 1322049209^4=\text{powerful}$
Let $a_1,a_2,a_3$ be integers.
The projective curve $X^4-Y^4=a_1^3 (a_1 X+a_2 Y)^2 Z^2$ is genus one elliptic curve and it may have infinitely many points $(X,Y,Z)$ and I believe it may have infinitely many solutions with $(X,Y)$ coprime.
Someone may try to find points on the elliptic curve for some $a_i$.
Comment appears to suggest that abc implies finitely many solution, but I believe abc doesn't imply this.
As pointedpointed by Peter Taylor, solutions to $X^4-Y^4=\text{powerful}$ exist like $10113607^4-4319999^4$, but in this particular case $X^2+Y^2$ is not powerful. It is a near miss and the only exponent of one is at the oddest prime $2$.
