22P+1 + 2P + 1 ≥ (2P + 1)²
=> 22P+1 + 2P + 1 = (2P + 1+k)²
=> 22P+1 + 2P + 1 = 2^2p + k²+2k+1+2*(2^p)*(k+1)
Up on simplication
=> 2^p(2^p-2k-1)=k(k+2)
if p>0, k should be even, let k=2c
=> 2^(p-2)(2^p-4c-1)=c(c+1)
(2^p-4c-1) < 4*2^(p-2)
=>
Case1:
Let c=2^(p-2)
=> 2^p-4c-1=-1 => not possible
Case2:
c+1=2^(p-2)
=> 2^p-4c-1 = 2^p-2^p+4-1=3
=> c=3
=> if c=3 => 2^(p-2)=4 => p=4
If p=0, 2^(2p+1)+2^p+1=4 which is a perfect square.

The only solutions are p=0,4