Why can’t mathematica find this residue?

You could use SeriesCoefficient instead:

SeriesCoefficient[(z+1)^2 Exp[3/z^2], {z, 0, -1}]

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Addendum

Another possibility is to note that the residue at 0 and the residue at infinity must sum to zero, since they are the only singularities of the function. Hence we can do:

- Residue[(z + 1)^2 Exp[3/z^2], {z, Infinity}]

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which is the same answer as before.

Or integrate around zero

Integrate[(z + 1)^2 Exp[3/z^2], {z, 1, I, -1, -I, 
    1}]/(2 Pi I) // Simplify

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