Slight correction. Your first bullet point, and hence your proof that ℚ cannot contain π, is correct.
A number x is algebraic over ℚ if and only if it generates a finite field extension, i.e. if x, x^2, x^3, etc. have a linear dependence relation.
However, as jopolous pointed out, you can get infinite dimensional algebraic field extensions by adjoining infinitely many algebraic numbers. For example, the set of all numbers which are algebraic over ℚ is a field, and this field is an infinite degree extension of ℚ.
A number x is algebraic over ℚ if and only if it generates a finite field extension, i.e. if x, x^2, x^3, etc. have a linear dependence relation.
However, as jopolous pointed out, you can get infinite dimensional algebraic field extensions by adjoining infinitely many algebraic numbers. For example, the set of all numbers which are algebraic over ℚ is a field, and this field is an infinite degree extension of ℚ.