First, we show that $3$ is primitive modulo $125$. This is easy, as $3$ is not a quadratic residue modulo $5$, so it is not modulo $125$. Now, we can note $3^5\equiv -7\pmod{125}$, so $3^{20}\equiv 2401\equiv 26\pmod{125}$, implying $3$ is primitive. Now, this implies there exists exactly one residue class modulo $100$, say $x\equiv m\pmod{100}$, such that \[3^x\equiv 22\pmod{125}\]Quick computations show \[3^{11}=3\cdot 3^{10}=3\cdot 49\equiv 22\pmod{125}\]Thus, it suffices to show \[3^{11+100m}+7\]has a large prime divisor. In particular, we'll show that \[101\mid 3^{11+100m}+7\]which follows from the fact $\varphi(101)=100$ and that \[3^{11}\equiv 3\cdot 41^2\equiv -7\pmod{101}\]as desired.