$$3^{n} - x^2 = 11$$

According to Silverman's method, we take the three cases $n=3a, n=3a+1,$ and $n=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.

$\bullet n=3a$
Let $X = 3^a, Y=x$, then we get $Y^2 = X^3 - 11.$
According to LMFDB, this elliptic curve has integer points $(X,Y)=(3,\pm 4), (15,\pm 58)$ with rank $2.$
Hence $(X,Y)= (3,\pm 4) \implies (n,x)=(3,\pm4).$

$\bullet n=3a+1$
Let $X = 3^{a+1}, Y=3x$, then we get $Y^2 = X^3 - 99.$
This elliptic curve has rank $0$ and has no integer point, so there is no integer solution $(n,x).$

$\bullet n=3a+2$
Let $X = 3^{a+2}, Y=9x$, then we get $Y^2 = X^3 - 891.$
This elliptic curve has integer points $(X,Y)=(31,\pm 170)$ with rank $1.$
Hence there is no integer solution $(n,x).$

Thus, there are only integer solutions $(n,x)=(3,\pm4).$

$$3^{n} - x^2 = 11$$

According to Silverman's answer, we take the three cases $n=3a, n=3a+1,$ and $n=3a+2.$
The problem can be reduced to finding the integer points on elliptic curves as follows.

$\bullet n=3a$
Let $X = 3^a, Y=x$, then we get $Y^2 = X^3 - 11.$
According to LMFDB, this elliptic curve has integer points $(X,Y)=(3,\pm 4), (15,\pm 58)$ with rank $2.$
Hence $(X,Y)= (3,\pm 4) \implies (n,x)=(3,\pm4).$

$\bullet n=3a+1$
Let $X = 3^{a+1}, Y=3x$, then we get $Y^2 = X^3 - 99.$
This elliptic curve has rank $0$ and has no integer point, so there is no integer solution $(n,x).$

$\bullet n=3a+2$
Let $X = 3^{a+2}, Y=9x$, then we get $Y^2 = X^3 - 891.$
This elliptic curve has integer points $(X,Y)=(31,\pm 170)$ with rank $1.$
Hence there is no integer solution $(n,x).$

Thus, there are only integer solutions $(n,x)=(3,\pm4).$