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 $(-1,\pm1)$, which don't result in roots for your equation.
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 $7^1+2=(\pm3)^2$.
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 your original equation $7^x+2=y^2$ has only the solutions $(x,y)=(1,\pm3)$.
