What methods might be appropriate to find the lowest integer bound for $t$ in the following inequality for a given $p$?
$(2t+2)! > 2^{p+1}$
The best I can do is by using the inverse gamma function:
$$ \begin{align} (2t+2)! & > 2^{p+1} \\ \Gamma(2t+3) & > 2^{p+1} \\ 2t+3 & > \Gamma^{-1}(2^{p+1}) \\ t & > \frac{\Gamma^{-1}(2^{p+1}) - 3}2 \end{align} $$
Giving the minimum bound on $t$ to be $\lceil{\frac{\Gamma^{-1}(2^{p+1}) - 3}2}\rceil$
David Cantrell gave a very good approximation of the inverse gamma function here. It is $$\Gamma^{-1}(x)=\frac{L(x)}{W\left(\frac{L(x)}{e}\right)}+\frac{1}{2}$$ using $$L(x)=\log \left(\frac{x+0.036534}{\sqrt{2 \pi }}\right)$$ $W(z)$ being Lambert function for which good approximations can be found here.
I hope and wish this could help you.