Solving for $n$

$\large \begin{aligned} \left(4^{n+7}\right)^3 & = \left(2^{n+23}\right)^4 \\ \left(2^{2n+14}\right)^3 & = \left(2^{n+23}\right)^4 \\ 2^{6n+42} & = 2^{4n+92} \\ \implies 6n+42 & = 4n + 92 \\ 2n & = 50 \\ \implies n & = \boxed{25} \end{aligned}$

$\large{\left(4^{n+7}\right)^3=\left(2^{n+23}\right)^4}$

From the power rule , $(a^n)^m=a^{n \times m}$ , so we have

$\large{4^{3(n+7)}=2^{4(n+23)}}$

$\large{2^{2(3)(n+7)}=2^{4(n+23)}}$

$\large{2(3)(n+7)=4(n+23)}$

$\large{6n+42=4n+92}$

$\large{2n=92-42}$

$\color{plum}\boxed{\large{n=25}}$