I think so too, but so have been unable to reason anything. Heading in brute-force is obviously the last resort for me, but was looking for something logical before I do that.
I believe the two squares immediately above and below the right 3 must be mines. Let me see if my reasoning makes sense:
The 4 has five remaining squares around it, but three of those overlap with the 3 to its left. That 3 already has one mine marked, so those three squares that neighbor both it and the 4 can have at most two mines in them. That means the two squares that don't overlap must both have mines.
I was more going for the right 3 having the two rightmost squares be free of mines. Because the 4 has five squares surrounding it and four of those five squares are also in contact with the 3 on the right, no matter how the mines are arranged all three mines would be to the left or center of the 3. This leaves the two right squares unable to be mined.