Calculating the exponential of an upper triangular matrix

You should decompose your matrix like this $$\begin{equation} \begin{pmatrix} 2 & 1 & 1 \\ 0 & 2 & 1\\ 0 & 0 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{pmatrix} + \begin{pmatrix} 0 & 1 & 1\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} \end{equation} $$ The left one commutes with the right one and the right one is nilpotent. So it is easy to compute.

We may use Cayley Hamilton theorem to find the exponential of $$ M=\begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}.$$

Note that the eigenvalues of $M$ are $1,1,1$

We write $$e^{tM} = \alpha (t)I+ \beta (t)M+ \gamma (t) M^2.$$

To find the coefficient functions we replace $M$ with $\lambda$ and solve the resulting system.

$$e^{t\lambda} = \alpha (t)I+ \beta (t)\lambda + \gamma (t) \lambda ^2.$$

Differentiating with respect to $\lambda $ we get $$te^{\lambda t}=\beta + 2\lambda \gamma$$ Differentiating again, we get $$t^2 e^{\lambda t }=2\gamma$$

With t=1 and $\lambda =1$ we get $$\alpha = e/2, \beta =0, \gamma = e/2$$

with these values we find $$e^M =e \begin{bmatrix} 1 & 1 & 3/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}. $$

Apply the same method with $\lambda =2$ we get $$e^N =e^2 \begin{bmatrix} 1 & 1 & 3/2\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{bmatrix}. $$

for the original matrix $$N=\begin{bmatrix} 2 & 1 & 1\\ 0 & 2 & 1\\ 0 & 0 & 2\\ \end{bmatrix}$$

Edit: As pointed out by @XTChen there is a much easier way to do this but I detail evaluation of the requested exponential nonetheless.

Note that $$\pmatrix{1&1&1\\0&1&1\\0&0&1}^n=\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}$$ \begin{align} \sum_{n=0}^\infty & \frac1{n!}\pmatrix{1&1&1\\0&1&1\\0&0&1}^n \\ &=\sum_{n=0}^\infty\frac1{n!}\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}\\ &=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\sum_{n=1}^\infty\pmatrix{1/n!&1/(n-1)!&(n+1)/(2(n-1)!)\\0&1/n!&1/(n-1)!\\0&0&1/n!}\\ &=\pmatrix{1&0&0\\0&1&0\\0&0&1}+\pmatrix{e-1&e&3e/2\\0&e-1&e\\0&0&e-1}\\ &=\pmatrix{e&e&3e/2\\0&e&e\\0&0&e}\\ &=\frac12e\pmatrix{2&2&3\\0&2&2\\0&0&2}\\ \end{align}