We shall consider 2 cases $y$ positive and $y$ negative. Obviously in either case $x \ge 0$. If $y$ is positive, firstly $5^x \cong y^4 +1 \cong 1(mod$ $2) \Longrightarrow y \cong 0(mod$ $2)$. So, we must have $5^x \cong 1(mod$ $8)$. Since $ord_8(5)=2$, $2|x$. Therefore $y^4 < (5^{(\frac{x}{2})})^2 = y^4 + 4y +1 < y^4 + 2y^2 +1 = (y^2 +1)^2$ if $y \ge 3$, contradiction. So, $y \le 2$. We get the solution $(x,y) =(2,2)$. If $y$ it is equivalent to find solutions of the equation $5^x = y^4 - 4y+1$ in positive integers. Using the exact same method as above, $(y^2 -1)^2 < (5^{(\frac{x}{2})})^2 < y^4$ for $y \ge 3$. So, $y\le 2$. But in this case, there are no solutions. The other case left is $y=0$ where we get the solution $(x,y)=(0,0)$

If $y=odd,$ then $RHS=even.$It is not possible. Then $y=even,$ then $RHS\equiv 1\pmod 8\to x=even\to$ $LHS$ is perfect square. Case $1:$ $y\geq 0\to (y^2+2)^2>y^4+4y+1>(y^2)^\to y^4+4y+1=(y^2+1)^2.$ Then $2y^2-4y=0.$ Then the solutions is $y=0,2.$ If $y=0\to x=0.$ If $y=2\to x=2.$ Case $2:$ $y<0$ then $a=-y$ where $a>0.$ Then $y^4+4y+1=a^4-4a+1.$ If $a=1,2$ don't have solution. If $a>2\to (a^2+1)^2>a^4-4a+1>(a^2-1)^2\to a^4-4a+1=a^4.$ Don't have solution.

we can prove x is odd when $y>2$. (just by looking to previous solutions). then $$5^x \equiv 2 \pmod 3 $$So $$y^4+ 4y+1 \equiv 2 \pmod 3 $$and there is no solution.

legendarybird wrote: we can prove x is odd when $y>2$. (just by looking to previous solutions). Please explain this Thank you @below

if x is even then LHS is a perfect square but if y>2 then ; $$ (y^2)^2<y^4+4y+1<(y^2+1)^2$$so RHS can't be a perfect square

If $x<0$ then $\text{LHS}\not\in\mathbb{Z}$ which is a contradiction. If $x=0$ we have that $1=1+4y+y^4\Longrightarrow y(y^3+4)=0$ thus either $y=0\text{ or }y=\sqrt[3]{-4}$ however the latter is not an integer and therefore is not a solution. So $(x,y)=(0,0)$ is one solution, now assume that $x\in\mathbb{Z}^+$ Notice that we can rewrite the equation as $5^x-1^x=y^4+4y\Longrightarrow\nu_2(5^x-1^x)=\nu_2(y(y^3+4))$ Furthermore by $\text{LTE}$ we obtain $\nu_2(5^x-1^x)=\nu_2(5-1)+\nu_2(x)\Longrightarrow\nu_2(x)+2=\nu_2(y(y^3+4))$ Moreover inspecting $\pmod 4$ yields $5^x\equiv1\pmod 4\text{ which forces }1+4y+y^4\equiv1\pmod 4\Longrightarrow y^4+4y\equiv y^4\equiv 0\pmod 4$ Thus $y\equiv0\text{ or }2\pmod 4\text{ and }y\equiv0\pmod 2$ so temporarily let $y^3=4k$ Now $\nu_2(x)+2=\nu_2(y)+\nu_2(y^3+4)=\nu_2(y)+\nu_2(4k+4)=\nu_2(y)+\nu_2(4)+\nu_2(k+1)=2+\nu_2(y)+\nu_2(k+1)$ Furthermore since $2\mid y$ we obtain $2+\nu_2(y)+\nu_2(k+1)\ge3$ therefore $\nu_2(x)+2\ge3\Longrightarrow\nu_2(x)\ge1$ thus $x$ is even. Let $x=2n$ Now the equation transforms into $(5^n)^2=1+4y+y^4$ Case 1: $y>0$ Notice that since $\text{LHS}$ is a perfect square, so must $\text{RHS}$ to be a perfect square. However notice that $(y^2)^2<y^4+4y+1<(y^2+1)^2$ for $y\ge3$ Thus $\text{RHS}$ is not a perfect square for $y\ge3$ which is a contradiction, therefore $y\le2$. Checking these cases yields only one solution, $(x,y)=(2,2)$ Case 2: $y<0$ The equation is equivalent to $(5^n)^2=y^4-4y+1$ However by the same process we obtain the following $(y^2-1)^2<y^4-4y+1<y^4$ for $y\ge3$ which is a contradiction, therefore $y\le2$. Checking the cases yields no solutions. So, to sum up $\boxed{(x,y)=(0,0)\text{ and }(2,2)\text{ are the only solutions for }x,y\in\mathbb{Z}}$ $\blacksquare$.