General term of ${(1+x)}^n$ is $T_{r+1}={}^n{\text{C}}_{r}x$

Fifth term of ${(1+x)}^n$ is $T_5=4.T_4$

$\Rightarrow {}^n{\text{C}}_4{x}^4=\left({}^n{\text{C}}_3{x}^3\right)4\cdots \left(1\right)$

And $T_4=6T_3$

$\Rightarrow {}^n{\text{C}}_3{x}^3=6\left({}^n{\text{C}}_2{x}^2\right)\cdots \left(2\right)$

$\begin{array}{l}\text{from\hspace{0.17em}\hspace{0.17em}}\left(1\right)\frac{{}^n{\text{C}}_4}{{}^n{\text{C}}_3}=\frac{4x^3}{x^4}\\ \Rightarrow \frac{n-4+1}{4}=\frac{4}{x}\\ \Rightarrow \frac{n-3}{4}=\frac{4}{x}\\ \Rightarrow x=\frac{16}{n-3}\cdots \left(3\right)\end{array}$

$\begin{array}{l}\text{from}\left(2\right)\frac{{}^n{\text{C}}_3}{{}^n{\text{C}}_2}=\frac{6x^2}{x^3}\\ \Rightarrow \frac{n-3+1}{3}=\frac{6}{x}\\ \Rightarrow \frac{n-2}{3}=\frac{6}{x}\\ x=\frac{18}{n-2}\cdots \left(4\right)\end{array}$

$\begin{array}{l}\text{from}\left(3\right)\&\left(4\right)\\ \frac{16}{n-3}=\frac{18}{n-2}\\ \Rightarrow 8(n-2)=9(n-3)\\ \Rightarrow 8n-16=9n-27\\ \Rightarrow n=11\end{array}$

Substitute $n=11$ in equation(3) we get $x=\frac{16}{11-3}=\frac{16}{8}=2$

$(x,n)=(2,11)$