<s>[INST] For every natural number $m$, $2^{2m}-1$ is divisible by $3$ Can anyone tell me if I missed any small details in this proof?
(basis) $2^{2(1)}-1 = 3*1$ which is divisible by 3. 
(induction) Fix $m>1$ so $p<m$. Hence for every natural number $p,$ $2^{2p}-1=3p.$ then for every natural number $(p+1),$ $2^{2(p+1)}-1=3(p+1)$ then
$2^{2(p+1)}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2*2^{2p}-1=3(p+1)$
$2(3p+1)-1=3(p+1)$ by induction hypothesis
$2(3p+1)+2^2-3=3(p+1)$
$6p+3=3(p+1)$
$3(p+1)=3(p+1)$
Thus, the statement is true for the natural number n whenever it is true for any natural number less than n.
 [/INST] $2^{2m} - 1 = (2^m)^2 - 1 = (2^m - 1)(2^m + 1); \tag 1$
of the three consecutive integers
$2^m - 1, \; 2^m, \; 2^m + 1 \tag 2$
precisely one of the must be divisible by $3$ (this is true of any three consecutive integers); clearly
$3 \not \mid 2^m; \tag 3$
therefore either
$3 \mid 2^m - 1 \; \text{or} \; 3 \mid 2^m + 1, \tag 4$
which implies
$3 \mid (2^m - 1)(2^m + 1) = 2^{2m} - 1. \tag 5$
Doesn't seem to use induction, does it?  But to be thorough, we need to prove that $3$ divides precisely one of any three consecutive positive integers . . . and induction might help here!  For example, restricting ourselves to the naturals $\Bbb N$, we see that $3$ divides precisely one out of $1, \; 2, \; 3$; now suppose for some $k$, $3$ divides precisely one of $k - 2$, $k - 1$, $k$; now consider the triplet $k - 1$, $k$, $k + 1$; if $3 \not \mid k - 2$, it must by hypothesis divide one of $k - 1$, $k$; but then $3 \not \mid k + 1$, lest it also divide $k - 2 = (k + 1) - 3$, which we have rejected.  So $3$ divides exactly one of the triplet $k - 1$, $k$, $k + 1$; and if $3 \mid k - 2$, we have $3 \mid (k - 2) + 3 = k + 1$, so once again $3$ is a factor of precisely one of $k - 1$, $k$, $k + 1$ and our little induction is done.
I find this method a little easier to get my head around, although I commend Bram28's suggestion for a proof, and his explanation, as well.
Note Added in Edit, Thursday 1 November 2018, 6:44 PM PST:  OK, so I fooled around with it a little more and found another proof which doesn't invoke the "three consecutive naturals" business, and I think is likely what Bram28 was getting at in his closing sentences; the argument performs induction directly $2^{2k} - 1$; suppose
$3 \mid 2^{2k} - 1; \tag 6$
then 
$\exists q \in \Bbb N, \; 2^{2k} - 1 = 3q; \tag 7$
thus,
$2^{2k} = 3q + 1; \tag 8$
now,
$2^{2(k + 1)} = 2^{2k + 2} = 2^{2k}2^2 = 4 \cdot 2^{2k} = 4(3q + 1) = 12q + 4, \tag 9$
whence
$2^{2(k + 1)} - 1 = 12q + 3 = 3(4q + 1), \tag{10}$
and that does it, does it not?  $OE\Delta$ End of Note.
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