<s>[INST] Find all integer solutions of the equation $x^2+15^a=2^b$ I found an integer solution for $$x^2+15^a=2^b$$ Which is $x=7$, when $a=1$ and $b=6$
In general: $x$ must be odd, $a$ must be odd and $b$ must be even
The problem is that I don't know wheater there are more solutions. Otherwise, how can I prove there is no other solution?
 [/INST] Another solution is $x=1,a=1,b=4$  
$b$ must be even for the right to be $1 \bmod 3$ as $x^2 \equiv 1 \pmod 3$.  Write $c=b/2$ and we are looking for $$(2^c-x)(2^c+x)=15^a$$  The two factors on the left are coprime, so we have the possibilities $$2^c-x=3^a\\2^c+x=5^a\\\text {and} \\2^c+x=15^a\\2^c-x=1$$
The solution $x=1,a=1,b=4,c=2$ is the first of these and $x=7,a=1,b=6,c=3$ is the second.  We cannot get any more from the second because we would have to have $2^{c+1}=15^a+1$ and the only perfect powers that differ by $1$ are $8,9$.  We cannot get any more from the first because we would have to have $2^{c+1}=5^a+3^a$ which can never be true $\bmod 16$ for $a \gt 1$
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