Let \(m>1\), \(r\geq 0\) be fixed non-negative integers and \(R\) be a ring. We say that a ring \(R\) has the property \(Q(m)\), if from \(m[x,y]=0\) it follows that \([x,y]=0\) for all \(x,y\in R\), where \([x,y]=xy-yx\) is the Lie commutator; the property \(P_1\), if for each \(x\in R\) there exists a polynomial \(f(X,Y)=f_x(X,Y)\in R\langle X,Y\rangle\) satisfying the condition that for all \(y\in R\), \(f(x,y)=f(x,y+1)=f(x,x+y)\) so that either \(y^r[x,y^m]=f(x,y)\) or \([x,y^m]y^r=f(x,y)\) holds for all \(y\in R\); the property \(P_2\), if for each \(x\in R\) there exist integers \(n=n(x)\geq 0\), \(p=p(x)\geq 0\) and \(q=q(x)\geq 0\) such that either \(y^r[x,y^m]=\pm x^p[x^n,y]x^q\) or \([x,y^m]y^r=\pm x^p[x^n,y]x^q\) for all \(y\in R\). The main results of the present paper are the following: Theorem 1. Let \(R\) be a ring with \(1\) and let \(m>1\), \(r\geq 0\). 1. If \(R\) satisfies the properties \(P_1\) and \(Q(m)\), then \(R\) is commutative. 2. If for every \(x\in R\) there exist integers \(n=n(x)>1\), \(p=p(x)\geq 0\) and \(q=q(x)\geq 0\) such that \(m\) and \(n\) are relatively prime and \(R\) satisfies the property \(P_2\), then \(R\) is commutative. Theorem 2. Let \(R\) be a ring such that \(x\in xR\) (or \(x\in Rx\), respectively) for each \(x\in R\) and let \(m>1\), \(r\geq 0\). Suppose that for every \(x\in R\) there exist integers \(n=n(x)>1\), \(p=p(x)\geq 0\) and \(q=q(x)\geq 0\) such that \(R\) satisfies the property \(P_2\). The ring \(R\) is commutative if one of the following conditions holds: 1. \(R\) has the property \(Q(m)\); 2. \(n>1\) and \(m>1\) are relatively prime integers.