to the equation How would on earth, anyone can prove this. frustrated! Please help.
 A: Note first that $b$ is odd.  Now look modulo $8$.  Since $b$ is odd, $b^2 \equiv 1 \pmod{8}$, so $b^2 - 5 \equiv 4 \pmod{8}$.  Thus, $b^2 -5$ is not divisible by 8, and in fact it is exactly divisible by $4$.  It follows that $a=2$, and we find $b = 3$.
A: Outline: If $a$ is odd, then $2^a\equiv \pm 2\pmod{5}$, so $2^a+5\equiv \pm 2\pmod{5}$, and therefore $2^a+5$ cannot be a square.  
So $a$ is even, say $a=2k$. We want $(2^k-b)(2^k+b)=-5$. But $-5$ has very few divisors.
A: $2^a+5=b^2$ so that $b$ is odd, i.e., $b=2k+1$ So $$ 2^a= 4k^2+4k-4$$
So $a=2+s$ so that $$ 2^s = k(k+1)-1 $$
$k(k+1)$ is even so that $$ s=0,\ k=1 $$ 
A: I think this may work, however I am not certain. 
Take the log base 2 of both sides of the equation then the resulting line will give all solutions to the problem. To prove that there aren't any other solutions first assume there is one solution (x,y) that doesn't lie on the line and then show that it must lie on the line and you have a contradiction. 
I would have preferred to leave this as a comment but I'm not allowed. 
I just noticed you want whole number solutions, but this method works for $x \in \Re$. You could, however, use something similar to show no other solutions than the ones you find exist. 
A: Just another order of the thoughts:
$$ 2^a + 4 = b^2-1 \qquad \to \qquad = (b+1)(b-1)$$
and it is immediate obvious that $b = 2b'+1$ must be odd (because the lhs is even).  Then 
$$ 4(2^{a-2} +1) = (2b'+2)(2b')=4(b'+1)b' $$
Cancel the factor $4$ 
$$ 2^{a-2} +1 = (b'+1)b' $$
Here the rhs is even, so the lhs must be even, and this is only possible, if $2^{a-2}=1$, so $a=2$ and we have the first equation numerically
$$ 4 + 4 = b^2-1 = 3^2-1 $$
