Twin primes whose sum is a cube $84537841287167$ and $84537841287169$ are a pair of twin primes.
The sum of $84537841287167+84537841287169$ is a cube.
Are there other examples of twin pairs $p$, $p+2$ whose sum is a cube?
Have they to have a particular form?
 A: To demonstrate that there are also huge solutions :
Define $$k=10^{100}+303593$$ $$s=4k^3-1$$ $$t=4k^3+1$$ then $(s,t)$ is a twin prime pair of the desired form which can be searched with this PARI/GP - routine 
? z=prod(j=1,3*10^4,prime(j));k=10^100-1;gef=0;while(gef==0,k=k+1;s=4*k^3-1;t=4*
k^3+1;if(gcd(s*t,z)==1,if(ispseudoprime(s)==1,print(k-10^100);if(ispseudoprime(t)
==1,gef=1))))

$\ s\ $and $\ t \ $ are proven primes with $\ 301\ $ digits. Assuming the generalized bunyakovsky conjecture, there are infinite many pairs of the desired form.
A: $(3,5),(107,109)$ and $(2634011,2634013)$ are only such twin prime pairs below $10^7$. You can check for more by increasing the range using this small Python code below. I just run it upto $10^8$, $(29659499,29659501)$ and $(57395627,57395629)$ are only such pairs with $10^7<p<10^8$. 

Edit: 
For optimization purpose, I will use the fact that all primes more than $3$ can be represented in the form $6k\pm 1$. 
First let check manually for $p=2,3$. For $p=2$ it's clearly not possible. For $p=3$, we have $3+5=8=2^3$, so, $(3,5)$ is such pair.
Suppose, $p=6k-1$, with $p>5$, then we have $12k=n^3$. That means, $12|n^3$. But, as $n^3$ is a perfect cube, and we have $12=3\cdot 2^2$, at least $3^3\cdot 2^3=216$ will divide $n^3$. So, the pair $(108m^3-1,108m^3+1)$ will be such pair if both of them are prime. 
While dealing with cubes, usually prefer to work in $\mathbb{Z}_7$, as any cube is either of $0,1,6$ in this field,i.e; $n^3\equiv 0,1,6\pmod{7}$ . So, we can have $2p+2\equiv 0\pmod{7}$, which gives $p\equiv 6\pmod{7}$, or, $2p+2\equiv 1\pmod{7}$ which implies $p\equiv 3\pmod{7}$ or $2p+2\equiv 6\pmod{7}$ which implies $p\equiv 2\pmod{7}$. Hence, only $3$ pssibilities. Here is the updated program:  
import math
import time

start_time=time.time()
def is_prime(n):
  flag=0
  if(n==2):
    return True
  if(n%2==0):
    return False
  else:
    for i in range(3,int(math.sqrt(n))+1,2):
      if(n%i==0):
        flag=1
        break
    if(flag==0):
      return True
    return False 

for i in range(1,1000):#change the number inside this braket to check for larger numbers
  c=(6*i)**3 
  p=c//2-1 
  if(p%7==2 or p%7==4 or p%7==6):
    if(is_prime(p) & is_prime(p+2)):
      print(p, "is such twin prime with sum",c) 

print(time.time()-start_time)

A: $p + (p + 2) = 2p + 2 = 2(p + 1)$
For the sum to be a cube, $p + 1$ must be divisible by 4, so that the sum becomes $2 \times 4m^3$ for some integer $m$. So $p + 1$ is of the form $4m^3$. 
