Prove $2^{135}+3^{133}<4^{108}$

First,

$$2^{135}=2^7\cdot2^{128}=2^7\cdot(2^8)^{16}<(2^8)^{17}=256^{17}\;.$$

Next,

$$3^{133}=3^3\cdot3^{130}=3^3\cdot(3^5)^{26}=3^3(243)^{26}<3^3(256)^{26}=27\cdot256^{26}\;.$$

Thus,

$$2^{135}+3^{133}<256^{17}+27\cdot256^{26}=(1+26\cdot256^9)256^{17}<256^{10}\cdot256^{17}=256^{27}=4^{108}\;.$$

While there are several more clever answers up there, I couldn't resist posting this answer.

2^135 =                          43556142965880123323311949751266331066368
3^133 =   2865014852390475710679572105323242035759805416923029389510561523
4^108 = 105312291668557186697918027683670432318895095400549111254310977536

So even by eye, you can confirm that $2^{135}+3^{133}<4^{108}$.

Using $3^5 < 2^8$ we have $3^{130}<2^{208}$.

$$2^{135}+3^{133}< 2^{135}+3^{3}2^{208}<2^{208}+3^{3}2^{208}=(1+27)4^{104}<4^4 \cdot 4^{104} $$