Finding all natural solutions to $2^a+5=b^2$ What are all possible natural number solutions $(a, b)$ to the equation

$$2^a+5=b^2 ?$$

The only solution I've found is 
$$2^2+5=3^2.$$
This may be the only solution but I don't have a proof.
 A: Hint If $a > 0$ then both sides are odd, in which case $b$ is odd, say, $2 m + 1$ for some natural $m$. Substituting this in the original equation, rearranging, and dividing by $4$ gives
$$2^{a - 2} + 1 = m^2 + m.$$

Additional hint The r.h.s. is $m(m + 1)$ and so is always even. For which $a$ is the l.h.s. even?

A: If we consider that $b^2-5=2^a$, with $a\ge3$ then it follows that $b^2$ must be odd so that the left hand side is even, and must satisfy $b^2-5\equiv0\pmod 8$ or $b^2\equiv5\pmod8$, that is, we need an odd perfect square congruent to $5\pmod8$.  However, odd squares can only be congruent to $1\pmod8$.
A: For the first equation we can see that $b$ is odd, since $2^a+5=b^2$ and $2^a$ is always even. So rewriting the equation you have $2^a+4= (b-1)(b+1)$. Is clear that if $a=1$, it doesn't has solution so $a\geq 2$. So
$$4(2^{a-2}+1)= (b-1)(b+1)$$
But $b$ is odd, so one between $b-1$ and $b+1$ is divided by 2 and the other by 4 (they're two consecutive even numbers), so the right side is divided by 8, and the only way for the left side to be divisible by 8 is when $2^{a-2}+1$ is even, that is when $2^{a-2}$ is odd, that only happens when $2^{a-2}=1$, so $a-2=0$, that lets $a=2$, that is the only solution you found.
