What is the next perfect square of the form 14444... in decimal notation?

144...4 is divisible by 4 hence it follows that 144...4 is a perfect square when 3611...1 is also a perfect square.

36 and 361 are special cases because others can be written by following

3611....111 = 4(25$m$ + 2) + 3 where $m$ is in $Z$

Consider the proof in the following question:

Proving that none of these elements 11, 111, 1111, 11111...can be a perfect square

Therefore, 3611...1 can not be a perfect square.

Brute force answer:

If $x^2 = 1\cdots4444$ then we can consider this mod $10000$ to see that we must have $x^2 \equiv 4444 \pmod{10000}$. To see if such $x$ exists, it is sufficient to consider $0 \le x < 10000$. The following C program terminates with no output, showing that no such $x$ exists.

#include <stdio.h>

int main(void) {
  int x;
  for (x = 0; x < 10000; x++) {
    if ((x * x) % 10000 == 4444) {
      printf("%d\n", x);
    }
  }
  return 0;
}

Note that you should run this on a system where int is at least 32 bits.