Ireland

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 $11^0\cdot 5^2-3^1\cdot 2^3=1$.