Then y^3 must be an odd number, and so:

y must be odd.  .....(i)

Now, p=2 gives y^3=16*2+1=33, which is not a perfect cube. Contradiction. 

Accordingly, p must be an odd prime. .....  (ii)

Now, 16p+1=y^3

=> 16p= y^3-1 = (y-1)(y^2+y+1)

From (i), we know that y-1 is an even number and, from (ii) we are au fait that p is odd. Since NO even number can divide an odd number, we must have: 

y-1 divides 16, and so y-1 correspond to all the divisors of 16 with the inclusion of 1 and itself. 

Accordingly,  y= 2,3,5,9,17

Substituting the above mentioned values for y in the equation:

y^3=16p+1, we observe that:

None of the values 2,3,5,9 yields a valid value for p.

But for y=17, we have:

p=(17^3-1)/16  = 4912/16 = 307

Consequently, 307 is the only possible value for p.