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Here is a technique that shows the commutator is bounded for $\alpha <1$. First note, as I observed in my comment, it is a bounded operator for $\alpha \le 0$. Now note that if $0<\eta<1$ and $A$ is a strictly positive operator then we have the identity $$A^\eta = c_\eta \int_0^\infty t^\eta \left (\frac{1}{t}- \frac{1}{t+A} -\frac{1}{t} \right ) dt$$ where $c_\eta \frac{1}{c_\eta} =\int_0^\infty t^\eta \left [\frac{1}{t}- \frac{1}{t+1} -\frac{1}{t} \right ]dt$ and the integral is to be understood in the strong sense (apply both sides to a vector in a dense core for the domain of $A^\eta$.) It follows that if $A$ and $B$ are two such operators then $$[A^\eta,B^\eta]=c_\eta^2 \int_0^\infty\int_0^\infty t^\eta s^\eta \left [ \frac{1}{t+A},\frac{1}{s+B} \right]ds dt.$$ The commutator of resolvents can be computed, for $A=1-\Delta$ and $B=1+x^2$ to give $$\left [\frac{1}{t+1-\Delta},\frac{1}{s+1+x^2} \right ] = -\frac{1}{t+1 -\Delta} \left [ 1-\Delta,\frac{1}{s+1 +x^2} \right ] \frac{1}{t+1 -\Delta}$$ $$= - 2 \frac{1}{t+1-\Delta} \left ( \frac{x}{(s+1+x^2)^2}\cdot \nabla+\nabla \cdot\frac{x}{(s+1+x^2)^2} \right ) \frac{1}{t+1-\Delta}.$$ The operator norm of the result is bounded by $$4 \frac{1}{(t+1)^{\frac{3}{2}}} \frac{1}{(s+1)^{\frac{3}{2}}}.$$ (To see this note that $\sup_x |x|/\sqrt{s+1 +x^2}= 1$. A similar computation on the Fourier side gives $\|\nabla/\sqrt{t+1-\Delta}\|=1$.) Plugging this into the integral representation gives $$\left \| \left [ (1-\Delta)^\frac{\alpha}{2}, (1+x^2)^\frac{\alpha}{2} \right ] \right \| \le \left (2 c_\eta \int_0^\infty \frac{t^\frac{\alpha}{2}}{(t+1)^{3/2}} dt \right )^2$$ which is finite if $0 <\alpha <1$.
Clearly this argument misses something since it doesn't give the boundary case $\alpha=1$, however I feel that a modification of this argument will show the commutator to be unbounded once $1<\alpha <2$ but I don't see the details. For $\alpha \ge 2$ I believe the commutator is unbounded and one can certainly show this, as I mentioned in my comment above, if $\alpha$ is an even natural number since the result is a partial differential operator.
Here is a technique that shows the commutator is bounded for $\alpha <1$. First note, as I observed in my comment, it is a bounded operator for $\alpha \le 0$. Now note that if $0<\eta<1$ and $A$ is a strictly positive operator then we have the identity $$A^\eta = c_\eta \int_0^\infty t^\eta \left ( \frac{1}{t+A} -\frac{1}{t} \right ) dt$$ where $c_\eta =\int_0^\infty t^\eta \left [ \frac{1}{t+1} -\frac{1}{t} \right ]dt$ and the integral is to be understood in the strong sense (apply both sides to a vector in a dense core for the domain of $A^\eta$.) It follows that if $A$ and $B$ are two such operators then $$[A^\eta,B^\eta]=c_\eta^2 \int_0^\infty\int_0^\infty t^\eta s^\eta \left [ \frac{1}{t+A},\frac{1}{s+B} \right]ds dt.$$ The commutator of resolvents can be computed, for $A=1-\Delta$ and $B=1+x^2$ to give $$\left [\frac{1}{t+1-\Delta},\frac{1}{s+1+x^2} \right ] = -\frac{1}{t+1 -\Delta} \left [ 1-\Delta,\frac{1}{s+1 +x^2} \right ] \frac{1}{t+1 -\Delta}$$ $$= - 2 \frac{1}{t+1-\Delta} \left ( \frac{x}{(s+1+x^2)^2}\cdot \nabla+\nabla \cdot\frac{x}{(s+1+x^2)^2} \right ) \frac{1}{t+1-\Delta}.$$ The operator norm of the result is bounded by $$4 \frac{1}{(t+1)^{\frac{3}{2}}} \frac{1}{(s+1)^{\frac{3}{2}}}.$$ (To see this note that $\sup_x |x|/\sqrt{s+1 +x^2}= 1$. A similar computation on the Fourier side gives $\|\nabla/\sqrt{t+1-\Delta}\|=1$.) Plugging this into the integral representation gives $$\left \| \left [ (1-\Delta)^\frac{\alpha}{2}, (1+x^2)^\frac{\alpha}{2} \right ] \right \| \le \left (2 c_\eta \int_0^\infty \frac{t^\frac{\alpha}{2}}{(t+1)^{3/2}} dt \right )^2$$ which is finite if $0 <\alpha <1$.
Clearly this argument misses something since it doesn't give the boundary case $\alpha=1$, however I feel that a modification of this argument will show the commutator to be unbounded once $1<\alpha <2$ but I don't see the details. For $\alpha \ge 2$ I believe the commutator is unbounded and one can certainly show this, as I mentioned in my comment above, if $\alpha$ is an even natural number since the result is a partial differential operator.