Add support for UUID fields

[brunomendola]

UUID fields are currently unsupported. Filtering by a UUID field within a JPA entity results in the following exception:

org.hibernate.query.SemanticException: Cannot compare left expression of type 'java.util.UUID' with right expression of type 'java.lang.String'
    at org.hibernate.query.sqm.internal.TypecheckUtil.assertComparable(TypecheckUtil.java:406) ~[hibernate-core-6.6.26.Final.jar:6.6.26.Final]
    at org.hibernate.query.sqm.tree.predicate.SqmComparisonPredicate.<init>(SqmComparisonPredicate.java:46) ~[hibernate-core-6.6.26.Final.jar:6.6.26.Final]
    at org.hibernate.query.sqm.tree.predicate.SqmComparisonPredicate.<init>(SqmComparisonPredicate.java:32) ~[hibernate-core-6.6.26.Final.jar:6.6.26.Final]
    at org.hibernate.query.sqm.internal.SqmCriteriaNodeBuilder.equal(SqmCriteriaNodeBuilder.java:2169) ~[hibernate-core-6.6.26.Final.jar:6.6.26.Final]
    at org.hibernate.query.sqm.internal.SqmCriteriaNodeBuilder.equal(SqmCriteriaNodeBuilder.java:192) ~[hibernate-core-6.6.26.Final.jar:6.6.26.Final]
    at io.github.queritylib.querity.jpa.JpaOperatorMapper.getEquals(JpaOperatorMapper.java:47) ~[querity-jpa-common-3.2.0.jar:na]
    at io.github.queritylib.querity.jpa.JpaOperatorMapper.getPredicate(JpaOperatorMapper.java:107) ~[querity-jpa-common-3.2.0.jar:na]
    at io.github.queritylib.querity.jpa.JpaSimpleCondition.toPredicate(JpaSimpleCondition.java:21) ~[querity-jpa-common-3.2.0.jar:na]
    at io.github.queritylib.querity.jpa.JpaQueryFactory.getPredicate(JpaQueryFactory.java:71) ~[querity-jpa-common-3.2.0.jar:na]
    at io.github.queritylib.querity.jpa.JpaQueryFactory.applyFilters(JpaQueryFactory.java:67) ~[querity-jpa-common-3.2.0.jar:na]
    at io.github.queritylib.querity.jpa.JpaQueryFactory.getJpaQuery(JpaQueryFactory.java:33) ~[querity-jpa-common-3.2.0.jar:na]
    at io.github.queritylib.querity.jpa.QuerityJpaImpl.findAll(QuerityJpaImpl.java:21) ~[querity-jpa-common-3.2.0.jar:na]