Find all integers $a, b, c, d \geq 1$ such that $4^{a} \cdot 5^{b}-3^{c} \cdot 11^{d}=1$.
We will show that the only solution is $(a, b, c, d)=(1,2,2,1)$.

Modulo 4, we have $3^{c+d} \equiv 3 \bmod 4$, which implies $c+d$ is odd.

Modulo 3, we find that $b$ is odd. Let's write $b=2 n$ with $n \geq 1$.

Modulo 8, using the fact that $c+d$ is odd and $b$ is even, we find that $4^{a} \equiv 4 \bmod 8$, which implies $a=1$.

Thus, $3^{c} \cdot 11^{d}=4 \cdot 5^{2 n}-1$, or equivalently $\left(2 \cdot 5^{n}-1\right)\left(2 \cdot 5^{n}+1\right)=3^{c} \cdot 11^{d}$. Since $2 \cdot 5^{n}-1$ and $2 \cdot 5^{n}+1$ are coprime, two cases arise.

Case 1. $2 \cdot 5^{n}-1=3^{c}$ and $2 \cdot 5^{n}+1=11^{d}$. The case $n=1$ gives the solution $c=2$ and $d=1$. We assume therefore $n \geq 2$ and $d \geq 2$. The order of 11 modulo 25 is 5, so $3\left|11^{20}-1\right| 11^{d}-1=2 \cdot 5^{n}$, which is absurd.

Case 2. $2 \cdot 5^{n}-1=11^{d}$ and $2 \cdot 5^{n}+1=3^{c}$. The case $n=1$ does not give any solutions, so we assume $n \geq 1$ and $c \geq 1$. The order of 3 modulo 25 is 20, so $3^{2}-1\left|3^{20}-1\right| 3^{c}-1=2 \cdot 5^{n}$, which is absurd.