Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve. 
$\sqrt{x+1} = \sqrt{x+2}$
$x+1 = x+2$
0 = 1
My question is, if there are two square roots on one side, then can I still square both sides in this way:
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$x+1 - (x+2) = x+3$
$x+1 - (x-2) = x+3$
$x = -4$
Or does squaring both sides cause something strange to happen on the left hand side?

No, what you did was wrong. 
this is because : 
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ does not imply $(x+1)-(x+2)=x+3$.
You need to square both sides, which means:
$(\sqrt{x+1} - \sqrt{x+2})^2=x+3$
$(x+1)+(x+2)-2\sqrt{(x+1)(x+2)}=x+3$.
As a sidenote, you can plug x=-4 into the original equation to check whether x=-4 is not a solution.
