We have $\sum_{j=1}^m j^2 = \frac{m(m+1)(2m+1)}6$. Hence, the question reduces to finding integral points on the elliptic curve: $$y^2 = \frac{x(x+1)(2x+1)}6.$$ Turning it to Weierstrass equation with integer coefficients, we get: $$(72y)^2 = (12x)^3 + 18 (12x)^2 + 72 (12x)$$

For many curves, integral points can be readily found by SAGE:

sage: EllipticCurve([0,18,0,72,0]).integral_points()
[(-12 : 0 : 1), (-9 : 9 : 1), (-8 : 8 : 1), (-6 : 0 : 1), (0 : 0 : 1), (6 : 36 : 1), (12 : 72 : 1), (288 : 5040 : 1)]

So, the only solutions are $m=\frac{0}{12}=0$, $m=\frac{12}{12}=1$, and $m=\frac{288}{12}=24$.

We have $\sum_{j=1}^m j^2 = \frac{m(m+1)(2m+1)}6$. Hence, the question reduces to finding integral points on the elliptic curve: $$y^2 = \frac{x(x+1)(2x+1)}6.$$ Turning it to Weierstrass equation with integer coefficients, we get: $$(72y)^2 = (12x)^3 + 18 (12x)^2 + 72 (12x).$$

For many curves, integral points can be readily found by SageMath, and in this case we are lucky:

sage: EllipticCurve([0,18,0,72,0]).integral_points()
[(-12 : 0 : 1), (-9 : 9 : 1), (-8 : 8 : 1), (-6 : 0 : 1), (0 : 0 : 1), (6 : 36 : 1), (12 : 72 : 1), (288 : 5040 : 1)]

So, the only solutions are $m=\frac{0}{12}=0$, $m=\frac{12}{12}=1$, and $m=\frac{288}{12}=24$.