How can I prove that $2^{\sqrt 7}$ is bigger than $5$?

Hint: $\sqrt{7}>\frac{5}{2}$.

Note that

$$5\lt2^\sqrt7\iff5^\sqrt7\lt(2^\sqrt7)^\sqrt7=2^7=128$$

But clearly $\sqrt7\lt3$, so $5^\sqrt7\lt5^3=125$.

Added later: Going beyond what the OP requests, here's a proof that $6\lt2^\sqrt7$, in enough detail that everything can be checked, I think, without resorting to a calculator.

Note first that

$$2\lt1+{7\over6}\lt\left(1+{1\over6}\right)^7=\left(7\over6\right)^7\implies 2\cdot6^7\lt7^7$$

and

$$3^5=243\lt256=2^8\implies3^5\cdot3^7\lt2^8\cdot3^7\implies3^{12}\lt2\cdot6^7$$

Putting these together, we have

$$3^{12}\lt7^7$$

From here, using the easy inequality $5/2\lt\sqrt7$ (which can be seen from $5^2=25\lt28=4\cdot7$) and the equality $(\sqrt7+1)(\sqrt7-1)=6$, we have

$$\begin{align} 3^{12}\lt7^7 &\implies3^6\lt7^{7/2}\\ &\implies3^6\lt7^{\sqrt7+1}\\ &\implies3^{(\sqrt7-1)(\sqrt7+1)}\lt7^{\sqrt7+1}\\ &\implies3^{\sqrt7-1}\lt7\\ &\implies3^{\sqrt7+1}\lt9\cdot7=63\lt64=2^6=2^{(\sqrt7-1)(\sqrt7+1)}\\ &\implies3\lt2^{\sqrt7-1}\\ &\implies6\lt2^\sqrt7 \end{align}$$

If there's an appreciably easier proof, I'd be interested to see one; I was a little surprised this one took as many steps as it did.

Added yet later: Here's an appreciably easier proof, using the fact that $7\lt64/9$ implies $\sqrt7\lt8/3$ and the inequality $3^8\lt2^{13}$, which can be verified by noting

$$3^8=81^2\lt90^2=8100\lt8160=8\cdot1020\lt8\cdot1024=2^{13}$$

Putting these together, we have

$$\begin{align} 3^8\lt2^{13} &\implies2^8\cdot3^8\lt2^8\cdot2^{13}\\ &\implies6^8\lt2^{21}\\ &\implies6^{8/3}\lt2^7\\ &\implies6^\sqrt7\lt2^7\\ &\implies6\lt2^\sqrt7 \end{align}$$

This solution is clearly worse, but it just goes to prove if you try enough stuff something will work.

You want to prove $a<b$, so prove $2^{(a^2)}<2^{(b^2)}$.

In this case we want $2^{(\log_2{5})^2}<2^7$

Of course $(\log_2{5})^2<3(\log_25)$ (since $2^3=8>5)$.

So it suffices to prove $(2^{log_2{5}})^3<2^7$. But of course $125<128$