Find $n$ and $m$, if $n,m$ are natural numbers, such that $m^6 + 279 = 2^n$. Suppose $n, m \in \mathbb{N}$. If $m^{6}+279=2^{n}$, find $n, m$. 
What are some ways to approach this question? Instinct told me to take logarithm but doesn't work to well.
 A: Suppose that $n$ is odd. Then
$$2=2*1^k\equiv 2*4^k=2^{2k+1}=2^n=m^6+279\equiv m^6=m^2*m^2*m^2(\text{mod }3)$$
If $m$ is divisible by $3$, then this is $0$ modulo $3$. If not, then
$$\equiv 1*1*1=1\ (\text{mod }3)$$
Since this is a contradiction either way, we conclude $n$ is even. Then we can simplify the equation to
$$m^6+279=4^k$$
for some $k\in\mathbb{N}$. Rearranging, we get
$$279=4^k-m^6=(2^k)^2-(m^3)^2=(2^k-m^3)(2^k+m^3)$$
Now, the prime factorization of $279$ is
$$279=3*3*31$$
This means that these prime numbers ($3$ twice and $31$ once) and only these prime numbers can divide
$$(2^k-m^3)(2^k+m^3)$$
Now, we have that
$$31>9=3*3$$
$$2^k+m^3>2^k-m^3$$
Thus, we know
$$31|(2^k+m^3)$$
However, since the other divisors are either $3$ or $9$, we may conclude
$$2^k+m^3\in\{31,63,279\}$$
Since $k>9$ and $m>7$ imply
$$2^k+m^3>279$$
we only need to check a finite number of combinations for possible solutions. In fact, we find one candidate:
$$2^2+3^3=31$$
where $k=2$ and $m=3$. If we plug these into the other factor, we get
$$2^k-m^3=2^2-3^3=4-27=-23$$
Since this is not divisible by $3$, we may conclude that the original equation has no solutions.
A: Based on achille hui's hint we find by exhaustive computation that
$$\forall m: \left(m^6 + 279\right) \bmod 7 \in \{0, 6\}$$
$$\forall n: 2^n \bmod 7 \in \{1, 2, 4\}$$
therefore the two expressions can never be equal.
