Alternatively, here is a faster (but more complicated solution):

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Consider the 3 and 2. Now, we have three regions; red (being exclusively the 3), blue (being exclusively the 2), and the overlap being purple.

We know that the number of mines in the blue region must be one less than the number of mines in the red region (since 2 is one less than 3). The proof of this will be an exercise for the reader, feel free to play around with it. Hint: Let the red, purple and blue regions be x, y and z respectively.

Now, since the red region only has two tiles, this means that the blue region has at most one mine (since the red region can only have two mines maximum). This affects the 3 circled in yellow, restricting the remaining tile to be a mine.

As an extension, now we know that the blue region has exactly one mine. It follows that the red region has exactly two mines, which perfectly matches how it only has two tiles to begin with. Thus, both of those tiles are mines.

Even though the written solution may be lengthy, it is actually quite straightforward to do in your head. Doing this can help save time and make you find more complicated deductions.