<s>[INST] Find the solutions in positive integers $m^2 + 615 =2^n$ Find  all positive integer solutions $m$ and $n$ of the equation
                     $$  m^2 + 615 =2^n.                   $$                                                                                                                                                                                                                                                        
 [/INST] Note that $3 \mid 615$ and $3$ is never a factor of $2^n$, so $3 \nmid m$.
Therefore $m^2 \equiv 1 \bmod 3$ and thus also $2^n\equiv 1 \bmod 3$.
Odd powers of $2$ are $\equiv 2 \bmod 3$ and even powers of $2$ are $\equiv 1 \bmod 3$, so we know that $n$ is even. So say $n=2k$.
Now we have $m^2 +615 =2^{2k}$ so $615= 2^{2k}-m^2 = (2^k)^2-m^2 = (2^k+m)(2^k-m)$
The factor pairs of $615$ are $\{(1,615),(3,205),(5,123),(15,41)\}$ so the possible values of $2^k$ from those are $\{308,104,64,28\}$. Only $64$ is a power of $2$ among these, so this gives $k=6$ and 
$\hspace{2in}\boxed{ n=12 \\ m=123-64=59}$
Check: $59^2+615 = 3481+615= 4096=2^{12} \quad\bigstar$
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