If you consider the Ljunggren-Nagell theorem "elementary" (see http://mathoverflow.net/q/206645/16510), then there is an elementary proof in your case. You ask about 
$$ 7^{2m} = 1 + 3y^2.$$
Clearly $y$ is even, say $y = 2u$. Then the equation can be rewritten in the form
$$\frac{7^m-1}{6} \cdot (7^m+1) = 2u^2.$$
Since $(7^m+1)-(7^m-1)=2$, the gcd of the two left-hand factors divides $2$. By unique factorization, each of the two left-hand factors is either a square or twice a square. 

If $m$ is odd, the first left-hand factor is odd. So 
$$ \frac{7^m-1}{6}= \square, \quad \text{which comparing with the last display forces}\quad 7^{m}+1 = 2\square. $$
If $m$ is even, both $(7^m-1)/6$ and $7^m+1$ are even. Moreover, $7^m+1 \equiv 2\pmod{4}$ --- hence, cannot be a square, so must be twice a square. Looking back at the displayed equation from the last paragraph, we see that again, $(7^m-1)/6=\square$. By Ljunggren's theorem, the only solutions to $(7^m-1)/6=\square$ are $m=1$ and $m=4$. But only $m=1$ is compatible with the condition $7^m+1=2\square$.