We consider four cases: $d=0$, $d=1$, $d=2$ and $d > 2$.

Case 1: $d=0$. In this case $3^c 2^d$ is odd, so there are no solutions.

Case 2: $d=1$. We have $11^{a} 5^{b} - 2 \cdot 3^{c} = 1$, so $c > 0$, and, reading mod 4, we see that $a$ is odd. Next reading mod 3, we get that $a+b$ is even. Hence $a = 2\alpha + 1$, $b = 2\beta + 1$, for some nonnegative integers $\alpha, \beta$. Then $55 \cdot 11^{2\alpha} 5^{2\beta} - 2 \cdot 3^c = 1$ and $c \ge 3$. Suppose that $c > 3$. Then we have $55(11^{2\alpha} 5^{2\beta} - 1) = 54(3^{c-3} - 1)$. Hence $3^{c-3} - 1$ is divisible by $11$. Since 5 is the order of 3 mod 11, $c-3 = 5\gamma$, for some nonnegative integer $\gamma$, and $3^{c-3} - 1$ is divisible by $3^5 - 1 = 2 \cdot 11^2$. This contradicts the fact that $11^2$ does not divide $55(11^{2\alpha} 5^{2\beta} - 1) = 54(3^{c-3} - 1)$. So, in this case $a=1$, $b=1$, $c=3$, $d=1$ is the only solution.

Case 3: $d=2$. Here $11^{a} 5^{b} - 4 \cdot 3^{c} = 1$, so reading mod 4, we see that $a$ is even. If $c=0$, then we have the solution $a=0$, $b=1$, $c=0$, $d=2$. Suppose that $c > 0$. Reading mod 3, we see that $a+b$ is even, so $a$ is even and $b$ is even. Write $a=2\alpha$, $b=2\beta$. Now we have $(11^{a} 5^{b} - 1)(11^{a} 5^{b} + 1) = 4 \cdot 3^c$ and this is impossible since the left-hand side is divisible by 8.

Case 4: $d > 2$. Reading $11^{a} 5^{b} - 3^{c} 2^{d} = 1$ mod 4, we see that $a$ is even. Reading mod 8, we find that $b$ is even and thus, $a=2\alpha$ and $b=2\beta$ for some nonnegative integers $\alpha, \beta$. Hence $(11^{a} 5^{b} - 1)(11^{a} 5^{b} + 1) = 3^{c} \cdot 2^{d}$. If $c=0$, then $(11^{a} 5^{b} - 1)(11^{a} 5^{b} + 1) = 2^{d}$. Since $\gcd(11^{a} 5^{b} - 1, 11^{a} 5^{b} + 1) = 2$, and $11^{a} 5^{b} - 1 > 2$, this is impossible. Hence $c > 0$ and we have two possibilities: (i) $11^{a} 5^{b} - 1 = 3^{c} \cdot 2$ and $11^{a} 5^{b} + 1 = 2^{d-1}$, and (ii) $11^{a} 5^{b} - 1 = 2^{d-1}$ and $11^{a} 5^{b} + 1 = 3^{c} \cdot 2$, where we again used $\gcd(11^{a} 5^{b} - 1, 11^{a} 5^{b} + 1) = 2$. In case (i), we obtain, by subtraction, $2^{d-1} = 3^{c} \cdot 2 + 2$ and thus $2^{d-2} = 3^{c} + 1$. So $2^{d-2} - 1 = 3^c$ and $d-2$ is even, say $d-2 = 2\delta$ and then $(2^d - 1)(2^d + 1) = 3^c$ and thus $\delta = 1$ and $c=1$, since $\gcd(2^d - 1, 2^d + 1) = 1$. But $11^{a} 5^{b} - 1 = 6$ has no solutions.
In case (ii), we have by subtraction $2 = 3^c \cdot 2 - 2^{d-1}$ and thus $2^{d-2} = 3^c - 1$. Then either $d=3$ and $c=1$ or, reading mod 4, $c=2\gamma$ is even. For $c > 1$, we have $2^{d-2} = (3^\gamma - 1)(3^\gamma + 1)$ and from $\gcd(3^\gamma - 1, 3^\gamma + 1) = 2$ we get $\gamma = 1$ and $d=5$. The case $c=1$ gives the solution $a=0$, $b=2$, $c=1$, $d=3$. The case $c=2$, $d=5$ does not yield solutions.

Combining all cases, we see that the solutions are $11^1 \cdot 5^1 - 3^3 \cdot 2^1 = 1$, $11^0 \cdot 5^1 - 3^0 \cdot 2^2 = 1$ and $11^0 \cdot 5^2 - 3^1 \cdot 2^3 = 1$.