jmc

Let $a=2^x$ and $b=3^y$. Substituting these values results in$$a^2-b^2=55$$Factor the difference of squares to get$$(a+b)(a-b)=55$$If $y<0$, then $55+3^{2y}<64$, so $y$ can not be negative. If $x<0$, then $2^{2x}<1$. Since $3^{2y}$ is always positive, the result would be way less than $55$, so $x$ can not be negative. Thus, $x$ and $y$ have to be nonnegative, so $a$ and $b$ are integers. Thus,$$a+b=55\text{ and }a-b=1$$$$\text{or}$$$$a+b=11\text{ and }a-b=5$$From the first case, $a=28$ and $b=27$. Since $2^x=28$ does not have an integral solution, the first case does not work. From the second case, $a=8$ and $b=3$. Thus, $x=3$ and $y=1$. Thus, there is only $\boxed{1}$ solution.