Noticed the prime number 97 is equal to the sum of the fourth power of 2 and the fourth power of 3

97 = 16+81=2^4+3^4

The equation is equivalent to

(x+2)^4+(x+1)^4=2^4+3^4

[(x+2)^4-2^4]+[(x+1)^4-3^4]=0

Use the difference of squares formula,

[(x+2)^2-2^2][(x+2)^2+2^2]+[(x+1)^2-3^2][(x+1)^2+3^2]=0

Continue to use the difference of squares formula,

x(x+4)(x^2+4x+8)+(x-2)(x+4)(x^2+2x+10)=0

Since there's a common factor x+4 in the both terms, we get the first solution

x=-4

Now let's continue for other solutions. Factor out the common factor as it will not be 0 for other solutions.

x(x^2+4x+8)+(x-2)(x^2+2x+10)=0

x^3+4x^2+8x+x^3+2x^2+10x-2x^2-4x-20=0

2x^3+4x^2+14x-20=0

x^3+2x^2+7x-10=0

Since x=1 is one of the solutions of the equation, we will factor the equation aiming to use x-1

x^3-1+2x^2-2+7x-7=0

(x-1)(x^2+x+1+2x+2+7)=0

(x-1)(x^2+3x+10)=0

Then we get

x=1

or

x=\dfrac{-3\pm\sqrt{9-40}}{2}=\dfrac{-3\pm i2\sqrt{10} }{2}

In summary, the equation has two real solutions and two imaginary solutios.