You need to make the following changes in your code. It should fix your issue

• No need for while (nums === Number) condition
• replace i + k == target with nums[i] + nums[k] == target. As i and k are indexes not actual values in array
• Don't use console.log("no result"); in else condition. Instead create a flag and check if it's false then do the console.log Otherwise it will console.log everytime nums[i] + nums[k] !== target.

const nums = [2, 7, 11, 15];
const target = 9;

let  flag = false;
function twoSum() {
  if (nums.length > 0) {
    for (let i = 0; i < nums.length -1; i++) {
      for (let k = 1; k < nums.length; k++) {
        if (nums[i] + nums[k] == target) {
        flag = true;
          console.log('indexes', [i, k], 'values', [nums[i], nums[k]]);
        }
      }
    }
    if (!flag) {
    console.log("No Result")
    }

  } else {
    console.log("Array is Empty");
  }
}

twoSum();

Assumed objective

Given an array of integers and a target, find any pairs of integers in the array whose sum matches the target.

Code Snippet

const nums = [2, 7, 11, 15];
const target = 9;

const twoSum = (tgt = target, arr = nums) => (
  arr.forEach(x => 
    arr.forEach(y =>
      x + y === tgt && console.log(x, y)
    )
  )
);

console.log('target = 9');
twoSum();

console.log('target = 4');
twoSum(4);

console.log('target = 30');
twoSum(30);

console.log('target = 55');
twoSum(100) || console.log('no matches');

Explanation

• Iterate through the array with each item taken as x
• Now, iterate again with each item taken as y
• If the sum of x and y matches target, then console.log

NOTES

• As seen in the test cases, same numbers may be considered twice. So, for target of 4 the code prints out 2 2.
• This may be avoided by changing the limits of the outer and inner loops.