I don't have a solution but since we're facing a narrow margins situation I'll answer. I followed the approach I guessed Gerhard Paseman's comment was talking about but there was an unforeseen complication.
If $\gcd(2^x-1,3^y-1)=1$ we have
$$2^x-1=p^2$$
$$3^y-1=2q^2$$
for integers $p,q$. We can factor the first one in $\mathbb{Z}[i]$
$$(-i)^x(1+i)^{2x}=(p+i)(p-i)$$
the two factors on the right hand side have the same modulus so they both must be a unit times $1+i$
$$p+i=u_1(1+i)^x$$
$$p-i=u_2(1+i)^x.$$
Then
$$2i=(u_1-u_2)(1+i)^x$$
we have unique factorization so the power of $1+i$ on the right hand side has to match the one on the LHS so $x\leq 2$. $x=2$ doesn't work mod 3 so we have $x=1$.
Now splitting into three cases for each residue of $y\mod 3$:
- $y=3w$
Then
$$w^3=2z^2+1$$
and to transform into an elliptic curve $w=w_1/2$ and $z=z_1/4$
$$z_1^2=w_1^3-8.$$
This is a Mordell curve so we can just look up the results. Bennet and Ghadermarzi have a table with all these solutions and looking up this and the other two cases, which turn out to have coefficients $-72$ and $-648$, we find that the three are the only solutions.
Now if $\gcd(2^x-1,3^y-1)\neq1$. I couldn't solve this case, but when $\gcd(x,y)=1$ I found some really weird behavior which made me think solving it in general might be harder than expected because it involves a relationship between $\text{ord}_p(2)$ and $\text{ord}_p(3)$. If a prime $p$ divides both sides of the gcd we have
$$2^x\equiv 3^y\equiv 1\mod p$$
$$\gcd(\text{ord}_p(2),\text{ord}_p(3)) = 1$$
This is relatively fast to calculate on a computer so I was able to find that $p=683,599479$ are the only possibilities under $10^9$. This is now OEIS sequence A344202, where you'll find 6 more terms which means mathematica has a faster way of calculating these than I do.
I feel like the $\gcd(x,y)\neq 1$ case might simplify to the previous one, but ruling out the few primes that satisfy the condition about the multiplicative orders of two and three will be difficult. For both of the primes I found we have $\text{ord}_p(2)\text{ord}_p(3)=p-1$ and there exists an unique element $r$ in $\mathbb{Z}_p$ so that $r^{\text{ord}_p(2)}\equiv 3$ and $r^{\text{ord}_p(3)}\equiv 2$. For $p=683$ it was $r\equiv 218$ and for $p=599479$, $r\equiv 45077$.