A general method, not necessarily best for this particular equation, is to split into three cases by writing $x=3u+v$ with $v\in\{0,1,2\}$. Then rewrite your equation as
$$ A(7^u)^3 + 2 = y^2\quad\text{with $A\in\{1,7,49\}$.} $$
So any solution to your equation gives an integer solution to one of the three equations
$$ w^3+2=y^2,\quad 7w^3+2=y^2,\quad49w^3+2=y^2.$$
There are then standard methods for finding integer points on these genus $1$ curves. 

**Addendum**: You can use the LMFDB to finish the solution. I'll rewrite the curves using the more typical $x,y$ variables. For $y^2=x^3+2$, we see from http://www.lmfdb.org/EllipticCurve/Q/1728/n/4 that the curve has rank $1$ and the only integer solutions are $(1,\pm2)$. For $7w^3+2=y^2$, we multiply by $7^2$ and change coordinates to get the elliptic curve $y^2=x^3+98$. Then http://www.lmfdb.org/EllipticCurve/Q/28224/dp/2 says that the rank is $1$, and the only integer points are $(7,\pm21)$, which gives the solution $7^1+2=3^2$ to your original equation. And for $49w^3+2=y^2$, we multiply by $7^4$ and change coordinates to get the elliptic curve $y^2=x^3+4802$. Then
http://www.lmfdb.org/EllipticCurve/Q/84672/fl/2 tells us that this curve has rank $0$ and no integer points. Hence you original equation $7^x+2=y^2$ has only the solutions $(x,y)=(1,\pm3)$.