Equation ax^p + by^q + cz^r = 0 is called super-fermat equation. 

If 1/p + 1/q + 1/r =1, the equation is called parabolic-type.

This type equation has infinitely many or finitely many solutions depending on a,b,c.([1].Cohen)

I show that the equation x^2 + y^4 = 5z^4 has infinitely many integer solutions.


Method 1.

x^2 + y^4 = 5z^4..........................................................(1)

First, we transform the equation (1) to the elliptic curve (2).

Let X=y/z, Y=x/z^2 then Y^2 = -X^4 + 5....................................(2)

Equation (2) is transformed to the next equation (3).

V^2=U^3+20U...............................................................(3)

By using Cremona's mwrank, we obtain the rank=1 and generator is P(4,-12) on (3).

Hence, (1) has infinitely rational solutions.

Small solutions(?) are shown below.
      x                       y             z
 P    2                       1             1
2P    358                     11            13 
3P    7297558                 2291          2005
4P    42286654776242          3014761       4398073 
5P    17100414992871911595602 133156642201  104958878089


Method 2.

We find a parametric solution.

Let x = gX^2+hX+p, y = X+q, z = X+r, p^2 + q^4 = 5r^4, then (1) becomes to (4).

(g^2-4)X^4+(4q-20r+2hg)X^3+(-30r^2+6q^2+2pg+h^2)X^2
+(-20r^3+2ph+4q^3)X+p^2+q^4-5r^4.........................................(4)

X = 4(qp^4-5rp^4+75r^5p^2-15r^3q^2p^2-250r^9+150r^6q^3-30r^3q^6-15q^3r^2p^2+3q^5p^2+2q^9)p^2
/(90r^2p^4q^2-600r^5p^2q^3+60r^2p^2q^6-300q^2p^2r^6+120q^5p^2r^3-225r^4p^4+1500r^8p^2-9q^4p^4-12q^8p^2+2000r^9q^3-600r^6q^6+80r^3q^9-2500r^12-4q^12+4p^6)

x = -p(-480q^15rp^4+1800000q^8p^2r^12-3600000q^5p^2r^15-5760q^17p^2r^3+72000q^14p^2r^6
    -300000r^15qp^4-640q^3p^10r+435000q^4p^4r^12-228000q^7p^4r^9+56400q^10p^4r^6-6720q^13p^4r^3
    -1400q^7p^8r-1440q^11p^6r-480000q^11p^2r^9+960r^3p^6q^9-4200000r^13p^4q^3+1410000r^10p^4q^6
    -108000r^7p^6q^5+17400r^4p^4q^12+240r^2p^10q^2+2400000r^11p^2q^9-9000000r^14p^2q^6+18000000r^17p^2q^3
    -1440r^2p^6q^10-700r^2p^8q^6-2160r^2p^4q^14+28800r^5p^2q^15-360000r^8p^2q^12+3000000q^2p^2r^18
    +180000r^10p^6q^2-960r^2p^2q^18+2240r^3p^10q-35000r^7p^8q-14600r^5p^8q^3+180000r^11p^6q
    -108000r^8p^6q^4+21600r^5p^6q^7+1850r^4p^8q^4+21600r^4p^6q^8-228000r^7p^4q^9-1350000r^14p^4q^2
    +1080000r^11p^4q^5-324000r^8p^4q^8+43200r^5p^4q^11-7900r^6p^8q^2+456000r^9p^6q^3-72000r^6p^6q^6
    +7480r^3p^8q^5+18750000r^24-16p^12-30000000r^21q^3+21000000r^18q^6-8400000r^15q^9+33600r^6q^18
    +312q^16p^4+359q^8p^8+384q^12p^6+152q^4p^10+192q^20p^2-3400r^4p^10+80375r^8p^8-840000r^12p^6
    +4875000r^16p^4-15000000r^20p^2-1920r^3q^21+2100000r^12q^12-336000r^9q^15+48q^24)

y = -1000p^2r^9+3p^4q^5-4p^2q^9-20p^6r+300p^4r^5+80r^3q^10-600r^6q^7+2000r^9q^4-2500qr^12
    -4q^13+30p^4q^3r^2+300p^2r^6q^3-60p^4r^3q^2+8p^6q+60r^2p^2q^7-600r^5p^2q^4-225qr^4p^4
    +1500qr^8p^2

z = 500p^2r^9+12p^4q^5+8p^2q^9-16p^6r+75p^4r^5-12rq^8p^2+120q^5p^2r^4-60p^4q^3r^2+80r^4q^9
   -600r^7q^6+2000r^10q^3-4rq^12-60p^2r^3q^6+30p^4r^3q^2+4p^6q-9rq^4p^4-300q^2p^2r^7-2500r^13

For example, substitute {p,q,r}={358,11,13} to {x,y,z}, then {x,y,z}={-17100414992871911595602, 133156642201, -104958878089}.
This case {p,q,r}={358,11,13} gives the same solution as the case 5P of method 1.




References

[1].Henri Cohen:Number Theory Volume 1:Tools and Diophantine Equations.